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 AX = 2e5 + 666;
const int MOD = 1e9 + 7;
long long quick(long long a, long long b) {
  long long ans = 1LL;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    b >>= 1;
    a = (a * a) % MOD;
  }
  return ans % MOD;
}
int p[AX];
long long cnt[AX] = {0LL};
long long pre[AX];
long long rear[AX];
int main() {
  int n;
  scanf("%d", &n);
  int x;
  int num = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (!cnt[x]) p[++num] = x;
    cnt[x]++;
  }
  pre[0] = 1LL;
  rear[num + 1] = 1LL;
  for (int i = 1; i <= num; i++) {
    pre[i] = (1LL * pre[i - 1] * (cnt[p[i]] + 1)) % (MOD - 1);
  }
  for (int i = num; i >= 1; i--) {
    rear[i] = (1LL * rear[i + 1] * (cnt[p[i]] + 1)) % (MOD - 1);
  }
  long long res = 1LL;
  for (int i = 1; i <= num; i++) {
    long long c = cnt[p[i]];
    if (!c) continue;
    long long sum = pre[i - 1] * rear[i + 1] % (MOD - 1);
    long long tmp = (1LL * c * (c + 1) / 2 % (MOD - 1) * sum) % (MOD - 1);
    res = (res * quick(p[i], tmp)) % MOD;
  }
  printf("%I64d\n", res);
  return 0;
}
