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

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.
#include <bits/stdc++.h>
using namespace std;
const int maxL = 200000, modd = 1000 * 1000 * 1000 + 7;
int N, M, G[maxL + 100];
int P[maxL + 10], S[maxL + 10];
int Ans;
int mul(int a, int x) {
  long long ret = 1, s = a;
  for (; x; x >>= 1) {
    if (x & 1) ret = ret * s % modd;
    s = s * s % modd;
  }
  return ret;
}
int main() {
  for (; scanf("%d", &N) != EOF;) {
    memset(G, 0, sizeof(G));
    for (int i = (1), _i = (N); i <= _i; i++) {
      int x;
      scanf("%d", &x);
      G[x]++;
    }
    P[0] = 1, S[maxL + 1] = 1;
    for (int i = (1), _i = (maxL); i <= _i; i++)
      P[i] = (long long)P[i - 1] * (G[i] + 1) % (modd - 1);
    for (int i = (maxL), _i = (1); i >= _i; i--)
      S[i] = (long long)S[i + 1] * (G[i] + 1) % (modd - 1);
    Ans = 1;
    for (int i = (1), _i = (maxL); i <= _i; i++)
      if (G[i]) {
        int x = (long long)G[i] * (G[i] + 1) / 2 % (modd - 1);
        x = (long long)x * P[i - 1] % (modd - 1);
        x = (long long)x * S[i + 1] % (modd - 1);
        Ans = (long long)Ans * mul(i, x) % modd;
      }
    printf("%d\n", Ans);
  }
  return 0;
}
