Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mx = 1e5 + 10;
long long mods[mx], a[mx];
long long Phi(long long x) {
  long long phi = 1;
  for (long long d = 2; d * d <= x; d++) {
    if (x % d != 0) continue;
    phi *= d - 1;
    x /= d;
    while (x % d == 0) {
      x /= d;
      phi *= d;
    }
  }
  if (x > 1) phi *= x - 1;
  return phi;
}
pair<long long, int> mult(pair<long long, int> x, pair<long long, int> y,
                          long long mod) {
  pair<long long, int> res = make_pair(x.first * y.first, x.second | y.second);
  if (res.first >= mod) {
    res.first %= mod;
    res.second = 1;
  }
  return res;
}
pair<long long, int> big_mod(pair<long long, int> x, long long p,
                             long long mod) {
  if (p == 0) return make_pair(1 % mod, 1 >= mod);
  if (p & 1) return mult(x, big_mod(x, p - 1, mod), mod);
  return big_mod(mult(x, x, mod), p / 2, mod);
}
pair<long long, int> bigmod(pair<long long, int> x, long long p,
                            long long mod) {
  if (p == 0) return make_pair(1 % mod, 1 >= mod);
  pair<long long, int> ans = make_pair(1 % mod, 1 >= mod);
  while (p) {
    if (p & 1) ans = mult(ans, x, mod);
    p = p >> 1;
    x = mult(x, x, mod);
  }
  return ans;
}
pair<long long, int> solve(int l, int r, int mid) {
  if (mods[mid] == 1) return make_pair(0, 1);
  if (l >= r) return make_pair(1, 0);
  pair<long long, int> pw = solve(l + 1, r, mid + 1);
  pair<long long, int> base = make_pair(a[l] % mods[mid], a[l] >= mods[mid]);
  long long pwr = pw.first + (pw.second ? mods[mid + 1] : 0);
  return bigmod(base, pwr, mods[mid]);
}
int main() {
  int n, q, l, r;
  scanf("%d", &n);
  scanf("%lld", &mods[0]);
  int id = 1;
  while (mods[id - 1] > 1) {
    mods[id] = Phi(mods[id - 1]);
    id++;
  }
  for (int i = (0); i < (n); i++) scanf("%lld", &a[i]);
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &l);
    scanf("%d", &r);
    l--;
    printf("%lld\n", solve(l, r, 0).first);
  }
  return 0;
}