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 MAX = 1e5 + 7;
int n, m, a[MAX], mods[200];
int phi(int x) {
  int aux = x;
  if (x % 2 == 0) {
    while (x % 2 == 0) x /= 2;
    aux /= 2;
  }
  for (int i = 3; i * i <= x; i += 2) {
    if (x % i == 0) {
      while (x % i == 0) x /= i;
      aux /= i;
      aux *= (i - 1);
    }
  }
  if (x > 1) {
    aux /= x;
    aux *= (x - 1);
  }
  return aux;
}
void pre() {
  mods[1] = m;
  int i = 2;
  while (mods[i - 1] > 1) {
    mods[i] = phi(mods[i - 1]);
    ++i;
  }
}
pair<int, int> mult(pair<int, int> a, pair<int, int> b, int mod) {
  pair<long long int, long long int> res =
      make_pair((1ll * a.first * b.first), (a.second | b.second));
  if (res.first >= mod) {
    res.first %= mod;
    res.second = 1;
  }
  return pair<int, int>(res);
}
pair<int, int> fpow(pair<int, int> b, long long int p, int mod) {
  if (p == 0) return make_pair((1 % mod), (1 >= mod));
  if (p & 1) return mult(b, fpow(b, p - 1, mod), mod);
  return fpow(mult(b, b, mod), p / 2, mod);
}
pair<int, int> solve(int l, int r, int ps) {
  if (mods[ps] == 1) return make_pair((0), (1));
  if (l == r) return make_pair((a[l] % mods[ps]), (a[l] >= mods[ps]));
  int ph = mods[ps + 1];
  pair<int, int> ex = solve(l + 1, r, ps + 1);
  return fpow(make_pair((a[l] % mods[ps]), (a[l] >= mods[ps])),
              ex.first + (ex.second ? ph : 0), mods[ps]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  pre();
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r, 1).first << '\n';
  }
  return 0;
}