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;
template <typename T1, typename T2>
inline void chkmin(T1 &first, T2 second) {
  if (first > second) first = second;
}
template <typename T1, typename T2>
inline void chkmax(T1 &first, T2 second) {
  if (first < second) first = second;
}
const string FILENAME = "input";
int n, m, q;
vector<int> res;
int arr[100001];
int func(int n) {
  vector<int> p;
  long double f = n;
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) p.push_back(i);
    while (n % i == 0) n /= i;
  }
  if (n != 1) p.push_back(n);
  for (int g : p) {
    long double k = 1;
    k = k - (k / g);
    f *= k;
  }
  return round(f);
}
inline int get_mod(long long a, int b) {
  if (a >= 32) {
    return 32 + ((a - 32) % b);
  }
  return a;
}
int powx(long long a, int b, int m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = get_mod(a * ans, m);
    a = get_mod(a * a, m);
    b >>= 1;
  }
  return ans;
}
int main() {
  srand(time(0));
  ios::sync_with_stdio(0);
  cin >> n >> m;
  res.push_back(m);
  while (m > 1) {
    m = func(m);
    res.push_back(m);
  }
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int L, R;
    cin >> L >> R;
    --L, --R;
    R = min(R + 1, L + (int)res.size()) - 1;
    int ans = 1;
    for (int j = R; j >= L; --j) {
      ans = powx(arr[j], ans, res[j - L]);
    }
    cout << ans % res[0] << '\n';
  }
}