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;
unordered_map<int, int> mp;
const int maxn = 2e5 + 233;
int phi(int n) {
  if (mp.find(n) != mp.end()) return mp[n];
  int res = n;
  int tmp = n;
  for (int i = 2; i * i <= tmp; i += 1) {
    if (tmp % i == 0) {
      res = res / i * (i - 1);
      while (tmp % i == 0) {
        tmp /= i;
      }
    }
  }
  if (tmp > 1) res = res / tmp * (tmp - 1);
  return mp[n] = res;
}
int a[maxn], n, m, q;
int mul(int x, int y, int mod) {
  return 1ll * x * y >= mod ? 1ll * x * y % mod + mod : 1ll * x * y % mod;
}
int pow_mod(int q, int w, int mod) {
  int ret = 1;
  while (w) {
    if (w & 1) ret = mul(ret, q, mod);
    q = mul(q, q, mod);
    w >>= 1;
  }
  return ret;
}
int solve(int l, int r, int mod) {
  if (mod == 1) {
    return 1;
  }
  if (l == r) {
    return mul(1, a[l], mod);
  }
  int cur_phi = phi(mod);
  int p = solve(l + 1, r, cur_phi);
  return pow_mod(a[l], p, mod);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i += 1) {
    cin >> a[i];
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r, m) % m << "\n";
  }
  return 0;
}