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 long long inf = 1000000000;
int n, m, w[100005], tot, a[105], nx[100005];
bool flag;
int ksm(int x, int y, int mo) {
  int ans = 1;
  bool f1 = 0;
  while (y) {
    if (y % 2) {
      flag += ((long long)ans * x > inf) | f1;
      if (flag)
        ans = (long long)ans * x % mo;
      else
        ans *= x;
    }
    f1 += ((long long)x * x > inf);
    if (f1 || flag)
      x = (long long)x * x % mo;
    else
      x *= x;
    y = y / 2;
  }
  return ans;
}
int phi(int n) {
  int result = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
int solve(int d, int l, int r) {
  int mo = a[d];
  if (l == r) {
    if (w[l] > inf) {
      flag = 1;
      return w[l] % mo;
    } else
      return w[l];
  }
  if (d == tot && r - l + 1 > 10) {
    flag = 1;
    return 0;
  }
  int phi = a[d + 1];
  int x = solve(min(tot, d + 1), l + 1, r);
  if (flag)
    x += phi;
  else if (x >= phi)
    x = x % phi + phi;
  return ksm(w[l], x, mo);
}
int main() {
  scanf("%d%d", &n, &m);
  a[0] = m;
  while (a[tot] > 1) tot++, a[tot] = phi(a[tot - 1]);
  a[tot + 1] = 1;
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  int q;
  scanf("%d", &q);
  nx[n + 1] = n + 1;
  for (int i = n; i >= 1; i--) nx[i] = (w[i] == 1) ? i : nx[i + 1];
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int tmp = nx[l];
    if (tmp <= r) r = tmp - 1;
    if (l > r)
      puts(m > 1 ? "1" : "0");
    else
      flag = 0, printf("%d\n", solve(0, l, r) % m);
  }
  return 0;
}