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 INF = 1e9 + 7;
vector<int> eum;
int w[100005];
int Euler(int num) {
  int p = 2, ans = 1;
  while (p * p <= num) {
    int ct = 1;
    while (num % p == 0) {
      ct *= p;
      num /= p;
    }
    if (ct > 1) ct = ct / p * (p - 1);
    ans *= ct;
    p++;
  }
  if (num > 1) ans *= num - 1;
  return ans;
}
long long power(int x, int t, long long m) {
  if (t == 0) return 1;
  long long cc = power(x, t >> 1, m);
  cc = cc * cc % m;
  if (t & 1) cc = cc * x % m;
  return cc;
}
int gcd(int m, int n) { return n == 0 ? m : gcd(n, m % n); }
int solve(int l, int r, int mid) {
  if (l > r) return 1;
  if (eum[mid] == 1) return 0;
  vector<int> cc;
  for (int k = l + 1; k <= r && k <= l + 4; k++) cc.push_back(w[k]);
  cc.push_back(1);
  for (int k = 0; k < (cc.size()); k++)
    if (cc[k] == 1) {
      long long cur = 1LL;
      for (int j = k - 1; j >= 0; j--) {
        if (cc[j] >= 30 || cc[j] >= 4 && cur >= 4) {
          cur = 30LL;
          break;
        }
        cur = power(cc[j], cur, 1LL << 62);
        if (cur > 30LL) {
          cur = 30LL;
          break;
        }
      }
      if (cur < 30LL)
        return power(w[l], cur, eum[mid]) % eum[mid];
      else
        return power(w[l], solve(l + 1, r, mid + 1) + eum[mid + 1], eum[mid]);
    }
}
int main() {
  int n, m, q, x, l, r;
  scanf("%d%d", &n, &m);
  eum.push_back(m);
  while ((x = eum.back()) > 1) eum.push_back(Euler(x));
  for (int k = 0; k < (n); k++) scanf("%d", &w[k]);
  scanf("%d", &q);
  for (int k = 0; k < (q); k++) {
    scanf("%d%d", &l, &r);
    l--;
    r--;
    printf("%d\n", solve(l, r, 0));
  }
  return 0;
}