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;
int n;
int d[200010];
int freq[200010];
const int MOD = 1000000007;
const int phiMOD = MOD - 1;
int L[200010], R[200010];
int ans = 1;
int LgPut(int baza, int exp) {
  int ret = 1;
  while (exp) {
    if (exp & 1) ret = (1LL * ret * baza) % MOD;
    exp >>= 1;
    baza = (1LL * baza * baza) % MOD;
  }
  return ret;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> d[i];
    ++freq[d[i]];
  }
  for (int i = 0; i <= 200001; ++i) L[i] = R[i] = 1;
  for (int i = 1; i <= 200000; ++i) {
    ++freq[i];
    L[i] = (1LL * L[i - 1] * freq[i]) % phiMOD;
  }
  for (int i = 200000; i >= 1; --i) {
    R[i] = (1LL * R[i + 1] * freq[i]) % phiMOD;
  }
  for (int i = 1; i <= 200000; ++i) {
    int x = freq[i] - 1;
    if (x == 0) continue;
    long long exponent = (1LL * L[i - 1] * R[i + 1]) % phiMOD;
    int nr;
    if (x % 2 == 0) {
      nr = x / 2;
      nr = (1LL * nr * (x + 1)) % phiMOD;
    } else {
      nr = (x + 1) / 2;
      nr = (1LL * nr * x) % phiMOD;
    }
    exponent = (1LL * exponent * nr) % phiMOD;
    ans = (1LL * ans * LgPut(i, exponent)) % MOD;
  }
  cout << ans;
  return 0;
}
