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 M = 1e9 + 7;
const int MAXN = 2e5 + 10;
long long cnt[MAXN], marray[MAXN], prefix[MAXN], suffix[MAXN];
long long fpow(long long n, long long k) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = (long long)r * n % M;
    n = (long long)n * n % M;
  }
  return r;
}
int main() {
  int m;
  scanf("%d", &m);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= m; i++) scanf("%lld", marray + i);
  for (int i = 1; i <= m; i++) cnt[marray[i]]++;
  prefix[0] = 1;
  suffix[MAXN - 1] = 1;
  for (int i = 1; i < MAXN; i++)
    prefix[i] = (prefix[i - 1] * (cnt[i] + 1)) % (M - 1);
  for (int i = MAXN - 2; i >= 2; i--)
    suffix[i] = (suffix[i + 1] * (cnt[i] + 1)) % (M - 1);
  long long ans = 1;
  for (int i = 1; i < MAXN; i++) {
    long long currentpow = 1LL * prefix[i - 1] * suffix[i + 1] % (M - 1);
    long long x = 1;
    for (int j = 1; j <= cnt[i]; j++) {
      x = (x * i) % M;
      ans = (ans * fpow(x, currentpow)) % M;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
