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 = 1e6;
const long long INF = 1e9 + 19;
const int MOD = 1e9 + 7;
int n;
int p[N];
int mx;
long long pref[N];
long long suf[N];
void read() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    mx = max(mx, x);
    p[x]++;
  }
}
long long rev(long long a, long long m) {
  if (a == 1) return 1;
  return ((1 - rev(m % a, a) * m) / a + m) % m;
}
long long binPow(long long a, long long b) {
  long long res = 1;
  for (; b > 0; b /= 2) {
    if (b % 2 == 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
  }
  return res;
}
void solve() {
  long long answer = 1;
  pref[0] = 1;
  suf[mx + 1] = 1;
  for (int i = 1; i <= mx; i++) pref[i] = pref[i - 1] * (p[i] + 1) % (MOD - 1);
  for (int i = mx; i >= 0; i--) suf[i] = suf[i + 1] * (p[i] + 1) % (MOD - 1);
  for (int i = 1; i <= mx; i++) {
    if (p[i] > 0) {
      long long tmp = pref[i - 1] * suf[i + 1] % (MOD - 1);
      answer = answer *
               binPow(i, tmp * (((p[i] + 1) * 1ll * p[i] / 2) % (MOD - 1)) %
                             (MOD - 1)) %
               MOD;
    }
  }
  cout << answer << endl;
}
void stress() {}
int main() {
  if (1) {
    int k = 1;
    for (int tt = 0; tt < k; tt++) {
      read();
      solve();
    }
  } else {
    stress();
  }
  return 0;
}
