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 MAXN = 2e5 + 10;
const int MAXP = 2e5;
const long long MOD = 1e9 + 7;
const long long K = MOD - 1;
int n = 0, m, te[MAXN];
long long p[MAXN], e[MAXN];
long long f[MAXN], g[MAXN];
void lucky() {
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int p;
    scanf("%d", &p), te[p]++;
  }
  for (int i = 2; i <= MAXP; i++) {
    if (!te[i]) continue;
    p[++n] = i, e[n] = te[i];
  }
}
long long qpow(long long a, long long b, long long MOD) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    b >>= 1, a = (a * a) % MOD;
  }
  return res;
}
int main() {
  lucky();
  f[0] = g[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = qpow(f[i - 1], e[i] + 1, MOD);
    long long t = ((((1 + e[i]) * e[i] / 2) % K) * g[i - 1]) % K;
    f[i] = (f[i] * qpow(p[i], t, MOD)) % MOD;
    g[i] = (g[i - 1] * (e[i] + 1)) % K;
  }
  printf("%d\n", int(f[n]));
  return 0;
}
