//a在模n乘法下的逆元，没有则返回-1
long long inv(long long a, long long n)
{
long long x, y;
long long t = exgcd(a, n, x, y);
if(t != 1) return -1;
return (x%n+n)%n;
}
int main()
{
int T, ca = 1;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &k, &ret);
int m = (int)sqrt(1.0 * MOD) + 10;
std::map<int, int> mp;
for(int i = 0; i < m; i++) {
int tmp = pow_mod(k, i);
if(mp.find(tmp) == mp.end()) {
mp[tmp] = i;
}
}
int ans = MOD;
for(int i = 0; i < m; i++) {
int tmp = pow_mod(pow_mod(k, i), m);
int x, y;
exgcd(tmp, MOD, x, y);
while(x < 0) x += MOD;
x = 1LL * x * ret % MOD;
y = 1LL * y * ret % MOD;
if(mp.find(x) != mp.end()) {
tmp = i * m + mp[x];
if(tmp < ans) {
ans = tmp;
}
}
}
printf("Case %d: %d\n", ca++, ans);
}
return 0;
}
long long inv(long long a)
{
long long x,y;
long long t=exgcd(a,M,x,y);
if(t!=1)
return -1;
return (x%M+M)%M;
}
long long inverse(long long num, long long mod)
{
long long x, y;
exgcd (num, mod, &x, &y);
while (x < 0)
x += mod, y -= num;
return x;
}
inline int exgcd(int a, int b, int &x, int &y) {//扩展Euclid
if(!b) {
x = 1; y = 0;
return a;
}
int ret = exgcd(b, a%b, y, x);
y -= a/b*x;
return ret;
}
static int64 inv(int64 x) {
int64 a, b;
exgcd(x, MOD, a, b);
if (a < 0)
a += MOD;
return a;
}
// X = r[i] (mod m[i]); 要求m[i]两两互质
// 引用返回通解X = re + k * mo;
void crt(ll r[], ll m[], ll n, ll &re, ll &mo) {
mo = 1, re = 0;
for (int i = 0; i < n; i++) mo *= m[i];
for (int i = 0; i < n; i++) {
ll x, y, d, tm = mo / m[i];
exgcd(tm, m[i], d, x, y);
re = (re + tm * x * r[i]) % mo;
} re = (re + mo) % mo;
}
void exgcd(__int64 a,__int64 b,__int64 &d,__int64&x,__int64 &y)
{
if(!b)
x=1,y=0,d=a;
else
{
exgcd(b,a%b,d,y,x),y-=x*(a/b);
}
}
static void exgcd(int64 a, int64 b, int64 &x, int64 &y) {
if (b == 0) {
x = 1;
y = 0;
} else {
int64 m, n;
exgcd(b, a % b, m, n);
x = n;
y = m - (a / b) * n;
}
}
pair exgcd(long long int a, long long int b){
if (b == 0){
pair p = {1,0};
return p;
} else {
long long int r = a / b;
long long int s = a % b;
pair x = exgcd(b, s);
pair p = {x.second, x.first - r * x.second};
return p;
}
}
long long exgcd(long long a, long long b, long long& x, long long &y)
{
if (b)
{
long long d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
x = 1;
y = 0;
return a;
}
int exgcd(int a, int b, int &x, int &y)
{
int gcd = a;
if(b == 0) {
x = 1;
y = 0;
} else {
gcd = exgcd(b, a % b, x, y);
x -= (a / b) * y;
std::swap(x, y);
}
return gcd;
}
LL china(int n,int *a,int *m){
LL M=1,d,x=0,y;
for(int i=0;i<n;i++)
M*=m[i];
for(int i=0;i<n;i++){
LL w=M/m[i];
d=exgcd(m[i],w,d,y);
y=(y%M+M)%M;
x=(x+y*w%M*a[i])%M;
}
while(x<0)x+=M;
return x;
}
long long exgcd(long long a,long long b,long long &x,long long &y)
{
if (b==0)
{
x=1;
y=0;
return a;
}
long long ans=exgcd(b,a%b,x,y);
long long temp=x;
x=y;
y=temp-(a/b)*y;
return ans;
}
lld exgcd(lld a,lld b,lld &x,lld &y){
if (!b){
x=1;
y=0;
return a;
}
else{
lld res=exgcd(b,a%b,x,y);
lld t=x;
x=y;
y=t-(a/b)*y;
return res;
}
}
// 返回值为 a 和 b 的最大公约数
int exgcd(int a, int b, int &x, int &y)
{
if (b) // 1
{
int r=exgcd(b,a%b,x,y);
int t=x; x=y; y=t-a/b*y;
return r;
}
else
{
x=1,y=0;
return a;
}
}
int main()
{
__int64 n=0,m,a,b,c,d,x0,y0,lcm;
int ca,i;
scanf("%d",&ca);
while(ca--)
{
scanf("%I64d",&m);
bool flag=1;
lcm=1;
for(i=1;i<=m;i++)
{
scanf("%I64d",&aa[i]);
lcm=lcm*aa[i]/gcd(lcm,aa[i]);
}
for(i=1;i<=m;i++)
{
scanf("%I64d",&rr[i]);
}
for(i=2;i<=m;i++)
{
a=aa[1],b=aa[i],c=rr[i]-rr[1];
exgcd(a,b,d,x0,y0);
if(c%d)
{
flag=0;
break;
}
__int64 t=b/d;
x0=(x0*(c/d)%t+t)%t;
rr[1]=aa[1]*x0+rr[1];
aa[1]=aa[1]*(aa[i]/d);
}
n++;
printf("Case %I64d: ",n);
if(!flag)
{
printf("-1\n");
continue;
}
__int64 ans=0;
if(rr[1]!=0)
{
printf("%I64d\n",rr[1]);
continue;
}
printf("%I64d\n",lcm);
}
}
// 如果有解则输出解并返回 true，否则返回 false
bool mod_equation(int a, int b, int n)
{
int x,y;
int d=exgcd(a,n,x,y);
if (b%d != 0)
return false;
else
{
int e = x*(b/d) % n;
for (int i=0; i<d; i++)
cout<< (e + i*(n/d)) % n <<endl;
return true;
}
}
lld solve(){
for (int i=1;i<=n;i++)
if (c[i]>=m[i]){
check=false;
return -1;
}
check=true;
lld ans=c[1],LCM=m[1],x,y,g,mm;
for (int i=2;i<=n;i++){
g=exgcd(LCM,m[i],x,y);
if ((c[i]-ans)%g) {check=false; return -1;}
ans=mod( ans + LCM * mod( (c[i]-ans)/g*x, m[i]/g ) , LCM/g*m[i]);
LCM=LCM/g*m[i];
}
return mod(ans,LCM);
}
int main()
{
int a, b, x, y;
printf("请输入第一个数: ");
scanf("%d", &a);
printf("请输入第二个数: ");
scanf("%d", &b);
//使用扩展欧几里德算法求最大公因数
//printf("%d 和 %d的最大公因数是: %d\n", a, b, gcd(a, b));
//使用扩展欧几里德算法计算a b的最大公因数
int maxComDivi = exgcd(a, b, &x, &y);
printf("\n%d 和 %d的最大公因数是: %d\n", a, b, maxComDivi);
printf("并且%d = %d * %d + %d * %d\n", maxComDivi, x, a, y, b);
}
long long exgcd (long long a, long long b, long long *x, long long *y)
{
if (b == 0)
{
*x = 1;
*y = 0;
return a;
}
else
{
long long d = exgcd (b, a % b, x, y);
long long t = *x;
*x = *y;
*y = t - a / b * (*y);
return d;
}
}
int exgcd(int a, int b, int * x, int * y)
{
int r; //最大公因数
int temp; //暂时保存x
if (b == 0)
{
*x = 1;
*y = 0;
return a; //a是最大公因数
}
r = exgcd(b, a % b, x, y); //x，y是一个指针
//x1 = y2, y1 = x2 - (a / b) * y2
temp = *x;
*x = *y;
*y = temp - (a / b) * (*y);
return r; //返回最大公因数
}
long long inv(long long a)
{
long long x, y;
exgcd(a, mod, x, y);
return (x % mod + mod) % mod;
}
// 利用exgcd求a在模m下的逆元，需要保证gcd(a, m) == 1.
ll inv(ll a, ll m) {
ll x, y, d; exgcd(a, m, d, x, y);
return d == 1 ? (x + m) % m : -1;
}
ll exgcd(ll a,ll b){
if(!b)return x=1,y=0,a;
ll d=exgcd(b,a%b),t=x;
return x=y,y=t-a/b*y,d;
}
ll rev(ll a,ll P){
exgcd(a,P);
while(x<0)x+=P;
return x%P;
}
int modMultInv(int R, int M){
long long int ret = exgcd(R, M).first;
if (ret < 0)
ret = -(-ret % M) + M;
return (int)(ret % 1000);
}
return (modPower(n,e-1,mod) * n) % mod;
}
unsigned long long int modulo(long long int n, long long int mod){
while (n < 0)
n += mod;
return n % mod;
}
pair exgcd(ulli a, ulli b){
if (b == 0){
pair p = {.x = 1, .y = 0};
return p;
} else {
pair m = {.x = a/b, .y = a%b};
pair n = exgcd(b, m.y);
pair r = {n.y, n.x - m.x * n.y};
return r;
}
}
ulli modMultInverse(ulli n, ulli mod){
return modulo(exgcd(n,mod).x, mod);
}
int main(void){
scanf("%llu %llu %llu", &n, &e, &c);
if (n % 2 == 0){
p = 2;
q = n / 2;
int Inval(int a,int b,int n){
int x,y,e;
exgcd(a,n,x,y);
e=(ll)x*b%n;
return e<0?e+n:e;
}
int inverse(int x, int mod) { // multiplicative inverse
int a = 0, b = 0;
if (exgcd(x, mod, a, b) != 1) return -1;
return (a % mod + mod) % mod; // C1: x & mod are co-prime
return fpow(x, mod - 2, mod); // C2: mod is prime
}
