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 N = 1e7 + 10;
const int mod = 1e9 + 7;
const double esp = 1e-12;
const unsigned long long base = 1331;
int phi[N], p[N], tot;
bool st[N];
void init() {
  phi[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!st[i]) phi[i] = i - 1, p[tot++] = i;
    for (int j = 0; j < tot && 1ll * i * p[j] < N; j++) {
      st[i * p[j]] = 1;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
}
long long qpow(long long x, long long y, long long p) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans = ans * x;
      if (ans >= p) ans = ans % p + p;
    }
    x = x * x;
    if (x >= mod) x = x % p + p;
    y >>= 1;
  }
  return ans;
}
vector<int> v;
long long s[N];
int Phi(int x) {
  int ans = x;
  for (int i = 0; i < tot && p[i] * p[i] <= x; i++) {
    if (x % p[i] == 0) {
      ans = ans / p[i] * (p[i] - 1);
      while (x % p[i] == 0) x /= p[i];
    }
  }
  if (x > 1) ans = ans / x * (x - 1);
  return ans;
}
void Init(int p) {
  while (p > 1) {
    v.push_back(p);
    if (p >= N)
      p = Phi(p);
    else
      p = phi[p];
  }
}
int main() {
  init();
  int n, m;
  cin >> n >> m;
  Init(m);
  v.push_back(1);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  int t, l, r;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    long long ans = 1;
    int k = min((int)v.size(), r - l + 1);
    while (k) {
      int id = l + k - 1;
      ans = qpow(s[id], ans, v[k - 1]);
      k--;
    }
    cout << ans % m << endl;
  }
}