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;
long long m[100005], n, q, mod;
vector<long long> mods;
long long eulerFunction(long long a) {
  long long k = a;
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      k -= k / i;
      while (a % i == 0) {
        a /= i;
      }
    }
  }
  if (a > 1) {
    k -= k / a;
  }
  return k;
}
long long f(long long x, long long y) { return x < y ? x : (x % y + y); }
long long binPow(long long x, long long y, long long tmod) {
  int res = 1;
  while (y) {
    if (y % 2 == 1) {
      res = f(res * x, tmod);
    }
    x = f(x * x, tmod);
    y >>= 1;
  }
  return res;
}
long long getResult(long long l, long long r, long long modPosition) {
  if (l > r || modPosition == mods.size()) {
    return 1;
  } else {
    long long tt = getResult(l + 1, r, modPosition + 1);
    long long tt2 = binPow(m[l], tt, mods[modPosition]);
    return tt2;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j, l, r, tr;
  cin >> n >> mod;
  for (i = 0; i < n; i++) {
    cin >> m[i];
  }
  cin >> q;
  mods.push_back(mod);
  while (mods.back() != 1) {
    mods.push_back(eulerFunction(mods.back()));
  }
  for (i = 0; i < q; i++) {
    cin >> l >> r;
    cout << getResult(l - 1, r - 1, 0) % mod << "\n";
  }
  return 0;
}