耗时好久，终于把这个题目给做出来了，不容易啊，弱校的数学始终是硬伤啊~~~
公式：
a^b=a^(b%phi(c)+phi(c))(mod c);
phi(c)为1~c中，与c互质的数的个数。
((a^b)^c)=a^(b^c)
a^(b^c)=a^(b^c%phi(c)+phi(c)) (mod c);
b^c%phi(c)=b^(c%phi(phi(c))+phi(phi(c)));
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<stdlib.h>
#define INF_MAX 0x7fffffff
#define INF 999999
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
struct node
{int u;int v;int w;bool friend operator < (node a, node b){return a.w < b.w;}
}edge[1001];
int gcd(int n,int m){if(n<m) swap(n,m);return n%m==0?m:gcd(m,n%m);}
int lcm(int n,int m){if(n<m) swap(n,m);return n/gcd(n,m)*m;}
int a[10001];
int b[10001];
int p[10001];
int phi(int x)
{int m;m=x;int i;for(i=2;i*i<=x;i++){if(x%i==0){m=m-m/i;while(x%i==0)x=x/i;}}if(x>1)m=m-m/x;return m;
}
int ex(int a,int b,int p)
{if(b==0)return 1%p;if(b==1)return a%p;int tx;tx=ex(a,b/2,p);tx=tx*tx%p;if(b%2==1)tx=tx*a%p;return tx;}
int main()
{int T,i,j;int m,n;cin>>T;while(T--){cin>>n>>m;for(i=0;i<n;i++){cin>>a[i];}if(a[0]>=m){puts("0");continue;}p[0]=m;for(i=1;i<n;i++){p[i]=phi(p[i-1]);}for(i=n-1;i>=1;i--){if(a[i]>=p[i])b[i]=p[i];else{int as=1;for(j=2;j<=a[i];j++){as*=j;as=as%p[i];}if(i==n-1){b[i]=as+p[i];}else{b[i]=ex(as,b[i+1],p[i])+p[i];}}}int as;as=1;for(i=2;i<=a[0];i++){as*=i;as=as%m;}cout<<ex(as,b[1],m)<<endl;}return 0;
}