Codeforces Round #338 (Div. 2) D 数学

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 p_i of size m, i.e. n = p₁·p₂·...·p_m. 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 p_i (2 ≤ p_i ≤ 200 000).

Output

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

Examples

Input

2
2 3

Output

36

Input

3
2 3 2

Output

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.

#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <string>
#include <complex>
#define ll __int64
#define mod 1000000007
using namespace std;
int m;
map<int,ll> mp;
ll a[200005];
ll quickpow(ll a,ll b)
{
    ll re=1;
    while(b>0)
    {
        if(b%2==1)
            re=(re*a)%mod;
        b/=2;
        a=(a*a)%mod;
    }
    return re%mod;
}
int main(){
    scanf("%d",&m);
    int jishu=0;
    ll exm;
    ll ans=1;
    for(int i=1; i<=m; i++){
        scanf("%I64d",&exm);
        if(mp[exm]==0){
            a[jishu++]=exm;
            }
        mp[exm]++;
    }
    ll re=1;
    for(int i=0;i<jishu;i++)
        re=re*(mp[a[i]]+1)%(2*(mod-1));
    for(int i=0;i<jishu;i++)
        ans=ans*quickpow(a[i],(re*mp[a[i]]/2)%(mod-1))%mod;
    printf("%I64d
",ans);
    return 0;
}