int main(){
int n;
//因此当模m有原根时，它有φ(φ(m))个原根。
while( scanf("%d", &n) != EOF ){
printf("%d\n", eular(eular(n)));
}
return 0;
}
int main() {
int m, k;
int i;
int total;
while(~scanf("%d%d", &m, &k)) {
if(m == 1)
printf("%d\n", k);
else {
long long res = eular(m);
long long cur = k / res;
if(k % res == 0)
cur--;
k = k - cur * res;
cur = cur * m + 1;
long long i;
long long temp;
for(i = cur; k != 0; i++) {
if(gcd(i, m) == 1) {
temp = i;
--k;
}
}
printf("%lld\n", temp);
}
}
}
// 根据摄像机方向改变billboard的纹理
bool BillboardSetObject::objectRendering(const MovableObject *mo, const Camera *camera)
{
const Quaternion &camQ = m_billboardSet->getCamQ();
Vector3 eular(MathEx::QuaternionToEular(camQ));
return true;
}
int main()
{
int n;
while(~scanf("%d", &n))
{
printf("%d\n", eular(n - 1));
}
return 0;
}
void eular(int u) {
used[u] = 1;
int i;
while (head[u] != -1) {
i = head[u];
head[u] = Next[head[u]];
if (vis[i])continue;
vis[i ^ 1] = 1;
eular(To[i]);
}
}
int main()
{
long long n;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
printf("%lld\n",eular(n+1));
}
return 0;
}
void solve(std::istream &in, std::ostream &out) {
int t;
in >> t;
while (t--) {
int n, m;
in >> n >> m;
init(n);
rep(i, n)
deg[i] = 0;
int u, v;
rep(i, m)
{
in >> u >> v;
u--, v--;
deg[u]++;
deg[v]++;
addedge(u, v);
addedge(v, u);
}
gao = -1;
rep(i, n)
{
if (deg[i] & 1) {
if (gao != -1) {
addedge(i, gao);
addedge(gao, i);
gao = -1;
} else gao = i;
}
}
rep(i, n)
used[i] = 0;
rep(i, n)
{
if (!used[i]) {
eular(i);
}
}
m <<= 1;
for (int i = 0; i < m; i += 2) {
if (!vis[i])out << 1 << endl;
else out << 0 << endl;
}
}
int print_eular(double x_0, double y_0, double x_n, double step)
{
double res = eular(x_0,y_0,x_n,step);
printf("The result of eular method is : %lf \n\r",res);
return 0;
}