Priests of the Quetzalcoatl cult want to build a tower to represent a power of their god. Tower is usually made of power-charged rocks. It is built with the help of rare magic by levitating the current top of tower and adding rocks at its bottom. If top, which is built from k - 1 rocks, possesses power p and we want to add the rock charged with power wk then value of power of a new tower will be {wk}p. 

Rocks are added from the last to the first. That is for sequence w1, ..., wm value of power will be

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After tower is built, its power may be extremely large. But still priests want to get some information about it, namely they want to know a number called cumulative power which is the true value of power taken modulo m. Priests have n rocks numbered from 1 to n. They ask you to calculate which value of cumulative power will the tower possess if they will build it from rocks numbered l, l + 1, ..., r. 

Input

First line of input contains two integers n (1 ≤ n ≤ 105) and m (1 ≤ m ≤ 109).

Second line of input contains n integers wk (1 ≤ wk ≤ 109) which is the power of rocks that priests have.

Third line of input contains single integer q (1 ≤ q ≤ 105) which is amount of queries from priests to you.

kth of next q lines contains two integers lk and rk (1 ≤ lk ≤ rk ≤ n). 

Output

Output q integers. k-th of them must be the amount of cumulative power the tower will have if is built from rocks lk, lk + 1, ..., rk.

Example

Input

6 1000000000
1 2 2 3 3 3
8
1 1
1 6
2 2
2 3
2 4
4 4
4 5
4 6


Output

1
1
2
4
256
3
27
597484987

Note

327 = 7625597484987
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const int N = 4e5 + 7;
const int M = 22;
const int multipleTest = 0;
int m, n;
int calcPhi(int m) {
  int res = m;
  for (int x = 2; x * x <= m; ++x) {
    if (m % x == 0) {
      while (m % x == 0) m /= x;
      res = res - res / x;
    }
  }
  if (m > 1) {
    res = res - res / m;
  }
  return res;
}
int w[N];
long long pw(long long val, long long pow, int mod) {
  long long res = 1;
  long long cur = val;
  while (pow) {
    if (pow & 1) {
      res = res * cur % mod;
    }
    cur = cur * cur % mod;
    pow >>= 1;
  }
  return res;
}
int gcd(int a, int b) { return (!a) ? b : gcd(b % a, a); }
vector<int> all;
long long calc(int u, int v, int j) {
  if (u == v) return min(w[u], w[u] % all[j] + w[u]);
  if (all[j] == 1) return 1;
  long long pwc = calc(u + 1, v, j + 1);
  if (w[u] == 1) return 1 % all[j];
  long long d = 1ll << 60, e = 1;
  for (long long t = 0; t < pwc; ++t) {
    d /= w[u];
    e *= w[u];
    if (!d) break;
  }
  if (!d) return pw(w[u], pwc, all[j]) + all[j];
  return min(e, e % all[j] + all[j]);
}
void solve() {
  cin >> n >> m;
  all.push_back(m);
  while (all.back() > 1) {
    all.push_back(calcPhi(all.back()));
  }
  for (int i = (1), _b = (n + 1); i < _b; ++i) scanf("%d", w + i);
  int q;
  cin >> q;
  while (q-- > 0) {
    int u, v;
    scanf("%d%d", &u, &v);
    int res = calc(u, v, 0);
    printf("%d\n", res % m);
  }
}
int main() {
  int Test = 1;
  if (multipleTest) cin >> Test;
  for (int i = 0; i < Test; ++i) {
    solve();
  }
}
