经典的欧拉降幂问题。
欧拉降幂公式：

n^x mod m = n^{φ(m)+(x mod φ(m))}

这个公式当且仅当x>φ(m)时成立。

//欧拉降幂公式
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define MOD(x,m) (x>m)?m+x%m:x
int n, m, a[20];
int phi[10000 + 10], kase;
void phi_table(int n)
{
    for (int i = 2; i <= n; i++)if (!phi[i])
        for (int j = i; j <= n; j += i) {
            if (!phi[j])phi[j] = j;
            phi[j] = phi[j] / i * (i - 1);
        }
}
LL fpow(LL a, LL m, LL mod)
{
    LL ret = 1;
    while (m) {
        if(m&1)ret = MOD(ret * a, mod);
        a = MOD(a * a, mod);
        m >>= 1;
    }
    return ret;
}
bool read()
{
    char ch;
    while (scanf("%c", &ch) && ch == '\n');
    if (ch == '#')return false;
    cin.putback(ch);
    scanf("%d%d", &m, &n);
    for (int i = 0; i < n; i++)scanf("%d", &a[i]);
    return true;
}
LL solve(int x, int mod)
{
    if (x + 1 == n || mod == 1)return MOD(a[x], mod);
    LL tmp = solve(x + 1, phi[mod]);
    return fpow(a[x], tmp, mod);
}
int main()
{
    phi_table(10000);
    while (read()) {
        printf("Case #%d: %lld\n",++kase, solve(0, m) % m);
    }
}