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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, a[100000 + 5], Q, p[100000 + 5], num = 0, R[100000 + 5],
                                           q[100000 + 5], top = 0;
int GetPhi(int x) {
  int res = x;
  for (int i = 2; i <= 100000; ++i)
    if (x % i == 0) {
      res = res / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
int pow(int x, int k, int mod) {
  int res = 1 % mod;
  for (; k; k >>= 1, x = 1LL * x * x % mod)
    if (k & 1) res = 1LL * res * x % mod;
  return res;
}
long long Mnpow(long long x, int k) {
  if (x == 1) return 1;
  long long res = 1;
  for (int j = 1; j <= k && res < 1e9; ++j) res = min((long long)1e9, res * x);
  return res;
}
int Solve(int l, int r, int m) {
  if (p[m] == 1) return 0;
  if (l == r) return a[l] % p[m];
  q[top = 1] = a[l + 1];
  for (int x = l + 2; top < 15 && x <= r && q[top] != 1; ++x) q[++top] = a[x];
  long long x = q[top];
  for (--top; top && x < p[m + 1]; --top) x = Mnpow(q[top], x);
  if (x < p[m + 1])
    return pow(a[l], x, p[m]);
  else
    return pow(a[l], Solve(l + 1, r, m + 1) + p[m + 1], p[m]);
}
int main() {
  n = read();
  p[num = 1] = read();
  for (; p[num] != 1;) ++num, p[num] = GetPhi(p[num - 1]);
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = n, lst = n + 1; i; --i) R[i] = lst, lst = (a[i] == 1) ? lst : i;
  Q = read();
  for (int i = 1; i <= Q; ++i) {
    int l = read(), r = read();
    printf("%d\n", Solve(l, r, 1));
  }
  return 0;
}
