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
#include <bits/stdc++.h>
const int maxn = 1e5 + 5;
const double EPS = 1e-6;
using namespace std;
map<long long, long long> mp;
long long a[maxn];
int mod(long long x, int p) { return x >= p ? x % p + p : x; }
long long kpow(long long a, long long b, long long p) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mod(ret * a, p);
    b >>= 1;
    a = mod(a * a, p);
  }
  return ret;
}
long long phi(long long k) {
  long long s = k, kk = k;
  if (mp.count(k)) return mp[k];
  for (long long 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);
  return mp[kk] = s;
}
long long solve(int l, int r, long long p) {
  if (p == 1) return a[l] % p + p;
  if (l == r) return mod(a[l], p);
  return kpow(a[l], solve(l + 1, r, phi(p)), p);
}
int main() {
  int n, q;
  long long p;
  scanf("%d%lld", &n, &p);
  for (int i = 1; i < (n + 1); ++i) scanf("%lld", &a[i]);
  scanf("%d", &q);
  int l, r;
  for (int i = 0; i < (q); ++i) {
    scanf("%d%d", &l, &r);
    printf("%lld\n", solve(l, r, p) % p);
  }
  return 0;
}
