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 = 1e5;
int rd() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - 48, ch = getchar();
  return x;
}
int n = 0, q = 0, mod = 0, cnt = 0;
int w[N + 5], u[N + 5], Pri[N + 5];
bool opt = false;
map<int, int> Phi;
void CalcPri() {
  int m = sqrt(mod);
  for (int i = 2; i <= m; i++) {
    if (!u[i]) Pri[++cnt] = i, u[i] = i;
    for (int j = 1; j <= cnt; j++) {
      if (Pri[j] > u[i] || Pri[j] > m / i) break;
      u[Pri[j] * i] = Pri[j];
    }
  }
}
int GetPhi(int x) {
  int res = x;
  for (int i = 1; i <= cnt; i++) {
    if (Pri[i] > x) break;
    if (x % Pri[i]) continue;
    res = res / Pri[i] * (Pri[i] - 1);
    while (x % Pri[i] == 0) x /= Pri[i];
  }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
void CalcPhi() {
  int m = mod;
  while (m > 1) Phi[m] = GetPhi(m), m = Phi[m];
}
int ksm(long long x, int k, int p) {
  long long res = 1;
  while (k) {
    if (k & 1) {
      res = res * x;
      if (res >= p) opt = true, res %= p;
    }
    x = x * x;
    if (x >= p) opt = true, x %= p;
    k >>= 1;
  }
  return res;
}
int Calc(int l, int r, int p) {
  if (l == r) return w[r] % p + (w[r] >= p) * p;
  if (p == 1) return 1;
  int x = Calc(l + 1, r, Phi[p]);
  opt = false;
  int res = ksm(w[l], x, p);
  return res + opt * p;
}
int main() {
  n = rd(), mod = rd();
  CalcPri(), CalcPhi();
  for (int i = 1; i <= n; i++) w[i] = rd();
  q = rd();
  for (int i = 1; i <= q; i++) {
    int l = rd(), r = rd();
    printf("%d\n", Calc(l, r, mod) % mod);
  }
  return 0;
}