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 N = 100005;
int n, m, a[N], phi[N];
int getphi(int mm) {
  if (mm == 1) return 0;
  int ph = 1;
  for (int x = 2; x * x <= mm; x++) {
    if (mm % x != 0) continue;
    int cnt = 0;
    while (mm % x == 0) {
      if (cnt == 0)
        ph *= (x - 1);
      else
        ph *= x;
      ++cnt;
      mm /= x;
    }
  }
  if (mm != 1) ph *= (mm - 1);
  return ph;
}
int power(int x, int v, int p) {
  int ans = 1;
  for (; v; v >>= 1, x = (long long)x * x % p)
    if (v & 1) ans = (long long)ans * x % p;
  return ans;
}
int work(int l, int r, int *c) {
  if (*c == 1) return 0;
  if (a[l] == 1) return 1;
  if (l == r) {
    return a[l] % (*c);
  }
  int aim = min(r, l + 7);
  for (int i = l + 1; i <= aim; ++i)
    if (a[i] == 1) {
      aim = i;
      break;
    }
  long long tot = a[aim];
  int flag = 0;
  for (int i = aim - 1; i >= l + 1; --i) {
    if (tot > 30 || a[i] >= *(c + 1)) {
      flag = 1;
      break;
    }
    long long ft = 1, xx = a[i];
    for (; tot; tot >>= 1, xx = xx * xx) {
      if (xx >= *(c + 1)) break;
      if (tot & 1) ft = ft * xx;
      if (ft >= *(c + 1)) break;
    }
    if (ft >= *(c + 1) || xx >= *(c + 1)) {
      flag = 1;
      break;
    }
    tot = ft;
  }
  if (flag)
    return power(a[l], work(l + 1, r, c + 1) + (*(c + 1)), *c);
  else
    return power(a[l], tot, *c);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  phi[0] = m;
  for (int i = 1; i <= 200; ++i) {
    phi[i] = getphi(phi[i - 1]);
    if (phi[i] == 0) break;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x == y) {
      printf("%d\n", a[x] % m);
    } else if (m == 1) {
      printf("%d\n", 0);
    } else {
      printf("%d\n", work(x, y, phi));
    }
  }
}