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 modulo(long long a, long long mod) {
  if (a < mod) {
    return a;
  } else {
    return (a % mod) + mod;
  }
}
long long mul(long long a, long long b, long long mod) {
  return modulo(a * b, mod);
}
long long expow(long long a, long long b, long long mod) {
  long long r = 1;
  while (b) {
    if (b & 1) {
      r = mul(r, a, mod);
    }
    a = mul(a, a, mod);
    b >>= 1;
  }
  return r;
}
bool prie(long long x) {
  if (x <= 1) {
    return 0;
  } else {
    for (long long d = 2; d * d <= x; d++) {
      if (x % d == 0) {
        return 0;
      }
    }
    return 1;
  }
}
vector<long long> primes;
long long phi(long long x) {
  long long sol = x;
  for (auto &d : primes) {
    if (d * d > x) break;
    bool is = 0;
    while (x % d == 0) {
      x /= d;
      is = 1;
    }
    if (is) {
      sol /= d;
      sol *= (d - 1);
    }
  }
  if (x > 1) {
    long long d = x;
    sol /= d;
    sol *= (d - 1);
  }
  return sol;
}
const long long N = (long long)1e5 + 7;
long long n, a[N];
vector<long long> mod;
long long solve(long long l, long long r, long long k) {
  if (k >= (long long)mod.size() || l > r) {
    return 1;
  }
  return expow(a[l], solve(l + 1, r, k + 1), mod[k]);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  long long ugar;
  cin >> ugar;
  mod.push_back(ugar);
  for (long long d = 2; d * d <= mod[0]; d++) {
    if (prie(d)) {
      primes.push_back(d);
    }
  }
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  while (mod.back() != 1) {
    mod.push_back(phi(mod.back()));
  }
  long long q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(l, r, 0) % ugar << "\n";
  }
}