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 a[100010];
map<long long, long long> mp;
long long mod(long long a, long long b) {
  if (a < b)
    return a;
  else
    return a % b + b;
}
long long quickpow(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = mod((res * a), m);
    a = mod(a * a, m);
    b = b >> 1;
  }
  return res;
}
long long euler(long long k) {
  long long i, s = k, x = k;
  if (mp.count(k)) return mp[x];
  for (i = 2; i * i <= k; i++) {
    if (k % i == 0) s = s / i * (i - 1);
    while (k % i == 0) k /= i;
  }
  if (k > 1) s = s / k * (k - 1);
  mp[x] = s;
  return s;
}
long long solve(int l, int r, long long m) {
  long long el = euler(m);
  if (m == 1ll || l == r) return mod(a[l], m);
  long long ans = solve(l + 1, r, el);
  return quickpow(a[l], ans, m);
}
int main() {
  int n, l, r;
  long long m, res, q;
  scanf("%d", &n);
  cin >> m;
  for (int i = 0; i < n; i++) scanf("%lld", a + i + 1);
  scanf("%lld", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    res = solve(l, r, m) % m;
    cout << res << endl;
  }
  return 0;
}