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 power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
long long a[100005], phi[100], m, findone[100005];
long long myphi(long long n) {
  long long result = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
void doIt() {
  phi[0] = m;
  phi[1] = myphi(m);
  for (long long i = 2; phi[i - 1] != 1; ++i) phi[i] = myphi(phi[i - 1]);
}
long long check(long long l, long long r) {
  r = min(r, findone[l]);
  long long ans = a[r];
  if (ans > 32) return ans;
  for (long long j = r - 1; j >= l; --j) {
    if (a[j] > 32 || ans > 32) return 33;
    long long z = 1;
    for (long long k = 1; k <= ans; ++k) {
      z *= a[j];
      if (z > 32) return 33;
    }
    ans = z;
  }
  return ans;
}
long long query(long long l, long long r, long long i) {
  long long ans = 1;
  if (l == r) return a[l] % phi[i];
  if (phi[i] == 1) return 0;
  if (check(l + 1, r) > 32)
    ans *= power(a[l], query(l + 1, r, i + 1) + phi[i + 1], phi[i]);
  else
    ans *= power(a[l], check(l + 1, r), phi[i]);
  return ans;
}
int _runtimeTerror_() {
  long long n, q, i, j;
  cin >> n >> m;
  doIt();
  for (i = 0; i < n; ++i) cin >> a[i];
  findone[n] = n;
  for (i = n - 1; i >= 0; --i) {
    if (a[i] == 1)
      findone[i] = i;
    else
      findone[i] = findone[i + 1];
  }
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << query(l, r, 0) << "\n";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}