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 maxn = 1e5 + 10;
long long pow_mod(long long p, long long k, long long mod) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * p;
    if (ans > mod) ans = mod + ans % mod;
    p = p * p;
    if (p > mod) p = mod + p % mod;
    k >>= 1;
  }
  return ans;
}
long long a[maxn], m[maxn];
int cnt = 0;
long long n, mm;
long long get_phi(long long x) {
  long long ans = 1;
  for (int i = 2; i * i <= x; i++) {
    if (x % i) continue;
    long long t = 1;
    while (x % i == 0) x /= i, t *= i;
    ans = ans * t / i * (i - 1);
  }
  if (x != 1) ans = ans * (x - 1);
  return ans;
}
void init() {
  m[0] = mm;
  long long tmp = mm;
  while (tmp > 1) {
    m[++cnt] = get_phi(tmp);
    tmp = m[cnt];
  }
}
long long dfs(int l, int r, int now) {
  if (m[now] == 1 || l == r) {
    long long ans = a[l] < m[now] ? a[l] : (m[now] + a[l] % m[now]);
    return a[l] < m[now] ? a[l] : (m[now] + a[l] % m[now]);
  }
  return pow_mod(a[l], dfs(l + 1, r, now + 1), m[now]);
}
int main() {
  scanf("%lld%lld", &n, &mm);
  init();
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  int k;
  scanf("%d", &k);
  while (k--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", dfs(l, r, 0) % mm);
  }
  return 0;
}
