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 N = 200005;
const long long MOD = 1e9 + 7;
int pw[N];
long long dp[N];
long long pref[N], suf[N];
long long binpow(long long a, long long p) {
  if (p == 0) return 1ll;
  long long x = binpow(a, p / 2ll);
  x = (x * x) % MOD;
  if (p % 2ll == 1ll) x = (x * a) % MOD;
  return x;
}
long long rev(long long n) { return (pref[n - 1] * suf[n + 1]) % (MOD - 1); }
int main() {
  dp[1] = 1ll;
  for (int i = 2; i < N; ++i) {
    dp[i] = (dp[i - 1] + (long long)i) % (MOD - 1);
  }
  int n;
  scanf("%d", &n);
  long long divcnt = 1;
  for (int i = 0; i < n; ++i) {
    int cur;
    scanf("%d", &cur);
    ++pw[cur];
  }
  pref[0] = 1ll;
  for (int i = 1; i < N; ++i) {
    pref[i] = (pref[i - 1] * (long long)(pw[i] + 1)) % (MOD - 1);
  }
  suf[N - 1] = 1ll;
  for (int i = N - 2; i >= 0; --i) {
    suf[i] = (suf[i + 1] * (long long)(pw[i] + 1)) % (MOD - 1);
  }
  long long ans = 1ll;
  for (int i = 1; i < N; ++i) {
    if (pw[i]) {
      long long curcnt = rev(i) % (MOD - 1);
      curcnt = (curcnt * dp[pw[i]]) % (MOD - 1);
      ans = (ans * binpow(i, curcnt)) % MOD;
    }
  }
  printf("%I64d", ans);
  return 0;
}
