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>
using namespace std;
const int N = 100005;
int n, p, Q, a[N], e[N], num = 0, cnt = 0, prime[N], phi[N];
bool vis[N];
inline int add(long long x, int mod) { return x >= mod ? x % mod + mod : x; }
int qm(int x, int k, int mod) {
  int sum = 1;
  while (k) {
    if (k & 1) sum = add(1ll * x * sum, mod);
    x = add(1ll * x * x, mod);
    k >>= 1;
  }
  return sum;
}
inline int getphi(int x) {
  if (x < N) return phi[x];
  int ret = x, lim = sqrt(x);
  for (int i = 1; i <= num && prime[i] <= lim && x > 1; i++)
    if (x % prime[i] == 0) {
      while (x % prime[i] == 0) x /= prime[i];
      ret = ret / prime[i] * (prime[i] - 1);
    }
  if (x > 1) ret = ret / x * (x - 1);
  return ret;
}
inline void priwork() {
  phi[1] = 1;
  for (int to, i = 2; i < N; i++) {
    if (!vis[i]) prime[++num] = i, phi[i] = i - 1;
    for (int j = 1; j <= num && i * prime[j] < N; j++) {
      vis[to = i * prime[j]] = 1;
      if (i % prime[j])
        phi[to] = phi[i] * (prime[j] - 1);
      else {
        phi[to] = phi[i] * prime[j];
        break;
      }
    }
  }
  e[++cnt] = p;
  while (p > 1) {
    e[++cnt] = getphi(p);
    p = e[cnt];
  }
  e[++cnt] = 1;
}
int main() {
  scanf("%d%d", &n, &p);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  priwork();
  scanf("%d", &Q);
  int l, r, ans;
  while (Q--) {
    scanf("%d%d", &l, &r);
    int j = min(cnt, r - l + 1);
    r = min(r, l + cnt - 1);
    ans = 1;
    for (int i = r; i >= l; i--, j--) ans = qm(a[i], ans, e[j]);
    printf("%d\n", ans % e[1]);
  }
  return 0;
}
