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>
const int64_t MOD = 1000000007ll;
int64_t mod_pow(int64_t a, int64_t b, int64_t m) {
  int64_t r = 1;
  while (b) {
    if (b & 1) {
      r = (r * a) % m;
    }
    a = (a * a) % m;
    b >>= 1;
  }
  return r;
}
int64_t sf[200001];
int64_t pf[200001];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int m;
  std::cin >> m;
  std::map<int, int> fac;
  for (int i = 0; i < m; ++i) {
    int p;
    std::cin >> p;
    fac[p]++;
  }
  int i = 0;
  for (auto&& pe : fac) {
    if (i > 0) {
      pf[i] = (pf[i - 1] * (1ll + pe.second)) % (MOD - 1);
    } else {
      pf[0] = (1ll + pe.second);
    }
    i++;
  }
  int size = i;
  i = size - 1;
  auto cont_end = fac.rend();
  for (auto it = fac.rbegin(); it != cont_end; ++it) {
    auto pe = *it;
    if (i == size - 1) {
      sf[i] = (1ll + pe.second);
    } else {
      sf[i] = (sf[i + 1] * (1ll + pe.second)) % (MOD - 1);
    }
    i--;
  }
  int64_t ans = 1;
  i = 0;
  for (auto&& pe : fac) {
    int64_t x = pe.second;
    x = (x * (x + 1) / 2) % (MOD - 1);
    if (i > 0) x = (x * pf[i - 1]) % (MOD - 1);
    if (i < size - 1) x = (x * sf[i + 1]) % (MOD - 1);
    ans *= mod_pow(pe.first, x, MOD);
    ans %= MOD;
    i++;
  }
  std::cout << ans << '\n';
  return 0;
}
