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 long long mod = 1000000007;
const int MAXN = 2e5 + 5;
long long dp[MAXN], cnt[MAXN];
long long binpow(int p, long long x) {
  if (x == 0) return 1ll;
  long long prv = binpow(p, x / 2);
  if (x & 1) {
    return p * ((prv * prv) % mod) % mod;
  } else
    return prv * prv % mod;
}
int n;
int a[MAXN], pw[MAXN];
int sz = 0;
int p[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d", &n);
  int prv = -1;
  for (auto i = 0; i != n; i++) scanf("%d", &p[i]);
  sort(p, p + n);
  for (auto i = 0; i != n; i++) {
    int x = p[i];
    if (prv != x) {
      a[sz] = x;
      pw[sz] = 0;
      sz++;
      prv = x;
    }
    pw[sz - 1]++;
  }
  dp[0] = cnt[0] = 1;
  for (auto i = 1; i != sz + 1; i++) {
    long long c = pw[i - 1] * 1ll * (pw[i - 1] + 1) / 2;
    c %= (mod - 1);
    c = c * cnt[i - 1] % (mod - 1);
    cnt[i] = cnt[i - 1] * (pw[i - 1] + 1) % (mod - 1);
    dp[i] = binpow(dp[i - 1], pw[i - 1] + 1);
    dp[i] = dp[i] * binpow(a[i - 1], c) % mod;
  }
  printf("%lld", dp[sz]);
  return 0;
}
