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 maxn = (int)1e5 + 1, maxm = 61, maxp = 31623, INF = 0x3f3f3f3f;
int tot, pr[maxp], d[maxp], n, m, mod[maxm], a[maxn], q, val[maxm];
int mod_pow(int x, int k, int p) {
  int ret = p > 1;
  for (; k > 0; k >>= 1, x = (long long)x * x % p)
    (k & 1) && (ret = (long long)ret * x % p);
  return ret;
}
int int_pow(int x, int k) {
  int ret = 1;
  for (; k > 0; k >>= 1, x = x * x) (k & 1) && (ret = ret * x);
  return ret;
}
int phi(int x) {
  int ret = x;
  for (int i = 0; pr[i] * pr[i] <= x; ++i)
    if (x % pr[i] == 0) {
      for (; x % pr[i] == 0; x /= pr[i])
        ;
      ret -= ret / pr[i];
    }
  if (x > 1) ret -= ret / x;
  return ret;
}
int main() {
  for (int i = 2; i < maxp; ++i) {
    if (!d[i]) pr[tot++] = d[i] = i;
    for (int j = 0, k; (k = i * pr[j]) < maxp; ++j) {
      d[k] = pr[j];
      if (d[i] == pr[j]) break;
    }
  }
  pr[tot] = maxp;
  scanf("%d%d", &n, mod + 0);
  for (m = 1; mod[m - 1] > 1; mod[m] = phi(mod[m - 1]), ++m)
    ;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  scanf("%d", &q);
  while (q--) {
    int L, R;
    scanf("%d%d", &L, &R);
    R = min(R, L + m - 1);
    bool chk = 0;
    int res = 1;
    for (int i = R - L; i >= 0; --i) {
      int ex = res + chk * mod[i + 1];
      if (pow(a[L + i], ex) < INF) {
        res = int_pow(a[L + i], ex);
        if (res >= mod[i]) {
          res %= mod[i];
          chk = 1;
        } else {
          chk = 0;
        }
      } else {
        res = mod_pow(a[L + i], ex, mod[i]);
        chk = 1;
      }
    }
    printf("%d\n", res);
  }
  return 0;
}