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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
long long mod(long long x, long long m) { return x >= m ? x % m + m : x; }
long long powmod(long long a, long long b, long long MOD) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = mod(ans * a, MOD);
    a = mod(a * a, MOD);
    b /= 2;
  }
  return ans;
}
long long m;
int n;
int q;
long long a[maxn];
map<long long, long long> vis;
long long euler(long long n) {
  if (vis.count(n)) {
    return vis[n];
  }
  long long ans = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans -= ans / i;
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans -= ans / n;
  vis[n] = ans;
  return ans;
}
long long solve(int l, int r, long long m) {
  if (l == r || m == 1) return mod(a[l], m);
  return powmod(a[l], solve(l + 1, r, euler(m)), m);
}
int main() {
  scanf("%d%lld", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  scanf("%d", &q);
  int l, r;
  while (q--) {
    scanf("%d %d", &l, &r);
    printf("%lld\n", solve(l, r, m) % m);
  }
  return 0;
}
inline void getInt(int* p) {
  char ch;
  do {
    ch = getchar();
  } while (ch == ' ' || ch == '\n');
  if (ch == '-') {
    *p = -(getchar() - '0');
    while ((ch = getchar()) >= '0' && ch <= '9') {
      *p = *p * 10 - ch + '0';
    }
  } else {
    *p = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') {
      *p = *p * 10 + ch - '0';
    }
  }
}