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 long long MX = 1e5 + 7, MOD = 1e9 + 7, INF = 1e9 + 7;
long long Phi(long long n) {
  long long ret = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i) continue;
    while (n % i == 0) n /= i;
    ret -= ret / i;
  }
  if (n > 1) ret -= ret / n;
  return ret;
}
long long Power(long long base, long long pow, long long mod) {
  long long ret = 1;
  while (pow) {
    if (pow & 1) ret = ret * base % mod;
    pow >>= 1;
    base = base * base % mod;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long W[n + 1], mod[100], Limit = 0;
  mod[0] = m;
  while (mod[Limit] > 1) mod[++Limit] = Phi(mod[Limit - 1]);
  vector<long long> Exp[MX];
  for (long long i = 2; i < MX; i++)
    for (long long x = 1; x < INF; x *= i) Exp[i].push_back(x);
  for (long long i = 1; i <= n; i++) cin >> W[i];
  long long q;
  cin >> q;
  while (q--) {
    long long L, R;
    cin >> L >> R;
    long long Ans = 1;
    for (long long i = min(R, L + Limit); i >= L; i--) {
      if (W[i] == 1)
        Ans = 1;
      else if (W[i] < MX && Exp[W[i]].size() > Ans)
        Ans = Exp[W[i]][Ans];
      else
        Ans = Power(W[i], Ans, mod[i - L]) + mod[i - L];
    }
    cout << Ans % m << "\n";
  }
}