计数原理

 

//cf 615D

D. Multipliers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ayrat has number n, represented as it's prime factorization pi of size m, i.e. n = p1·p2·...·pm. Ayrat got secret information that that the product of all divisors of n taken modulo 109 + 7 is the password to the secret data base. Now he wants to calculate this value.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of primes in factorization of n.

The second line contains m primes numbers pi (2 ≤ pi ≤ 200 000).

Output

Print one integer — the product of all divisors of n modulo 109 + 7.

Examples

input

Copy

2
2 3

output

Copy

36

input

Copy

3
2 3 2

output

Copy

1728

Note

In the first sample n = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.

In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <cstdlib>
typedef long long ll;
#define lowbit(x) (x&(-x))
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
using namespace std;
const int   mod= 1e9+7;
const int N=2e5+9;
int num[N],a[N];
ll l[N],r[N];
int n,x;
ll poww(ll a,ll b,ll mod){
    ll ret=1%mod;
    while(b){
        if(b&1ll) ret=ret*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return  ret%mod;
}
int main()
{
    scanf("%d",&n);
    int   top=0;
    for(int  i =0;i< n;i++){
        scanf("%d",&x);
        if(!num[x]) a[++top] =x;//共top个数
        num[x]++;
    }
    /*
    l[i]:1~i个因子的组合方案数
    r[i]:i~top个因子的组合方案数
    */
    l[0]=r[top+1]=1;//自己不存在，但可以和其他的因子组合
    for(int i=1;i<=top;i++){
        l[i]=l[i-1]*(1ll+num[a[i]])%(mod-1);
        /*
        第i个因子可以取或不取：
        不取 ：就是l[i-1]
        取：l[i-1]*num[a[i]]
        */
    }
    for(int i=top;i>=1;i--){
        r[i]=r[i+1]*(1ll+num[a[i]])%(mod-1);
    }
    ll ans=1;
    for(int i=1;i<=top;i++){
        ll numm=num[a[i]];
        numm=(numm*(numm+1))/2%(mod-1);
        //1+2+3+……+numm
        //1个，2个....(a[i])分别和前后组合的结果拼接
        ll z =l[i-1]*r[i+1]%(mod-1);//前后的组合方案数
        numm=numm*z%(mod-1);//a[i]出现了numm次
        ans=ans*poww(a[i],numm,mod)%mod;
    }
    printf("%lld
",ans);
    return    0;
}