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

<image>

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;
long long get_phi(long long x) {
  long long ans = x;
  if (x == 1) return 0;
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans /= i, ans *= (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans /= x, ans *= (x - 1);
  return ans;
}
int n, m, q;
long long num[100001];
long long phi[51];
long long my_pow(long long a, long long b, long long c) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (1ll * ret * a) % c;
    b >>= 1, a = (1ll * a * a) % c;
  }
  return ret;
}
bool check(int l, int r, long long value) {
  if (l == r) return num[r] == value;
  if (num[l] == 1) return value == 1;
  double k = log(value) / log(num[l]);
  long long d = (long long)floor(k + 0.5);
  if (abs(k - d) >= 1e-7)
    return false;
  else
    return check(l + 1, r, d);
}
pair<long long, bool> query(int l, int r, int depth) {
  long long val = 1;
  if (l == r)
    val = num[r] % phi[depth];
  else if (phi[depth] == 1)
    val = 0;
  else {
    pair<long long, bool> right = query(l + 1, r, depth + 1);
    if (right.second)
      val = my_pow(num[l], right.first, phi[depth]);
    else
      val = my_pow(num[l], right.first + phi[depth + 1], phi[depth]);
  }
  bool equal = check(l, r, val);
  return make_pair(val, equal);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &num[i]);
  phi[0] = m;
  for (int i = 1;; i++) {
    phi[i] = get_phi(phi[i - 1]);
    if (!phi[i]) break;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", query(l, r, 0).first);
  }
  return 0;
}