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 mod = 1e9 + 7;
const int phi = mod - 1;
const int maxn = 2e5 + 10;
int pre[maxn], suf[maxn];
map<int, int> mp;
int fac[maxn][2];
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int x;
    scanf("%d", &x);
    ++mp[x];
    pre[i] = suf[i] = 1;
  }
  pre[m] = suf[m] = 1;
  pre[m + 1] = suf[m + 1] = 1;
  int cnt = 0;
  for (auto &it : mp) {
    fac[++cnt][0] = it.first;
    fac[cnt][1] = it.second;
  }
  for (int i = 1; i <= cnt; ++i) {
    pre[i] = (long long)pre[i - 1] * (fac[i][1] + 1) % phi;
  }
  for (int i = cnt; i >= 1; --i) {
    suf[i] = (long long)suf[i + 1] * (fac[i][1] + 1) % phi;
  }
  long long ans = 1;
  for (int i = 1; i <= cnt; ++i) {
    long long tmp = (long long)pre[i - 1] * suf[i + 1] % phi;
    tmp = tmp * ((long long)fac[i][1] * (fac[i][1] + 1) / 2 % phi) % phi;
    tmp = (tmp + phi);
    ans = ans * qpow(fac[i][0], tmp) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
