int main()
{
int i, j, k;
while ( scanf("%d %d", &CASH, &N) != -1 )
{
if( N == 0)
{
printf("%d\n", 0);
continue;
}
memset(dp, 0, sizeof(dp));
sum_bills = 0;
int t_num, t_val;
for(i=1; i<=N; i++)
{
scanf("%d %d", &t_num, &t_val);
if( t_num > 0 )
{
int alfa = 1;
while( (t_num - alfa) >= 0 )
{
t_num -= alfa;
val[++sum_bills] = alfa * t_val;
alfa *= 2;
}
if ( t_num )
val[++sum_bills] = t_num * t_val;
}
}
if( CASH == 0 )
{
printf("%d\n", 0);
continue;
}
get_ans();
for(i=CASH; i>=0; i--)
{
if( dp[i] == 1 )
{
printf("%d\n", i);
break;
}
}
}
return 0;
}
int get_ans(int s,int k)//返回s的容量从第k件物品开始放最少需要几件
{
int i,ans=99999999,t;
if(s%w[k]==0)//若能整除直接返回
return s/w[k];
if(k==N)//没有放满且后面没有物品就返回-1
return -1;
int l=min(s/w[k],w[k+1]-1);//遍历的上限
//比如说接下来的数是w[k+1]，我只要考虑w[k]最多少放(w[k+1]-1)次即可
for(i=0;i<=l;i++)//遍历少放w[k]的次数0至w[k+1]-1
{
t=get_ans(s%w[k]+i*w[k],k+1);
if(t==-1)//不能放满
continue;
ans=min(t+s/w[k]-i,ans);
}
return ans;
}
int main()
{
sum = 0;
memset(dp, 0, sizeof(dp));
scanf("%d %d", &N, &B);
int i;
for(i=1; i<=N; i++)
{
scanf("%d", &heights[i]);
sum += heights[i];
}
get_ans();
for(i=B; i<=sum; i++)
{
if( dp[i] == 1 )
{
printf("%d\n", (i-B));
break;
}
}
return 0;
}
