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
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int Mo[N], a[N];
inline int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') ch = getchar(), o = -1;
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
inline int phi(int x) {
  int ret = x, y = x;
  for (int i = 2; i * i <= y; i++)
    if (x % i == 0) {
      while (x % i == 0) x /= i;
      ret = ret / i * (i - 1);
    }
  if (x > 1) ret = ret / x * (x - 1);
  return ret;
}
inline void calc(int &x, int y, int M) {
  long long c = 1LL * x * y;
  x = (c < M ? c : c % M + M);
}
inline int Pow(int x, int y, int M) {
  int ret = 1;
  while (y) {
    if (y & 1) calc(ret, x, M);
    calc(x, x, M), y >>= 1;
  }
  return ret;
}
inline int solve(int l, int r, int t) {
  if (Mo[t] == 1) return 1;
  if (l == r) {
    if (a[l] < Mo[t]) return a[l];
    return a[l] % Mo[t] + Mo[t];
  }
  return Pow(a[l], solve(l + 1, r, t + 1), Mo[t]);
}
int main() {
  int n, m, q;
  cin >> n >> m;
  Mo[1] = m;
  for (int i = 1; Mo[i] ^ 1;) Mo[++i] = m = phi(m);
  for (int i = 1; i <= n; i++) a[i] = gi();
  cin >> q;
  while (q--) {
    int l = gi(), r = gi();
    printf("%d\n", solve(l, r, 1) % Mo[1]);
  }
  return 0;
}
