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 read() {
  bool f = 0;
  long long x = 0;
  char c = getchar();
  while (!isdigit(c)) f = (c == '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return !f ? x : -x;
}
void print(int x) {
  if (x < 10) {
    if (x < 0)
      putchar('-'), x = -x;
    else {
      putchar('0' + x);
      return;
    }
  }
  print(x / 10);
  putchar('0' + x % 10);
  return;
}
const int MAXN = 200000;
const int INF = 0x3f3f3f3f;
long long Mul(long long x, long long y, long long Mod) {
  x *= y;
  return x >= Mod ? x % Mod + Mod : x;
}
long long Pow(long long x, long long y, long long Mod) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = Mul(ret, x, Mod);
    x = Mul(x, x, Mod), y >>= 1;
  }
  return ret;
}
int cnt, p[35], a[MAXN + 5];
long long Mi(int L, int R, int i) {
  if (p[i] == 1 || L == R) return Mul(1, a[L], p[i]);
  return Pow(a[L], Mi(L + 1, R, i + 1), p[i]);
}
bool pc[MAXN + 5];
int pcnt, pr[MAXN + 5];
void Prime() {
  for (int i = 2; i <= MAXN; i++) {
    if (!pc[i]) pr[++pcnt] = i;
    for (int j = 1; i * pr[j] <= MAXN; j++) {
      pc[i * pr[j]] = 1;
      if (i % pr[j] == 0) break;
    }
  }
  return;
}
int phi(int n) {
  int ret = n;
  for (int i = 1; i <= pcnt && pr[i] * pr[i] <= n; i++)
    if (n % pr[i] == 0) {
      ret = ret / pr[i] * (pr[i] - 1);
      while (n % pr[i] == 0) n /= pr[i];
    }
  if (n != 1) ret = ret / n * (n - 1);
  return ret;
}
int main() {
  Prime();
  int n = read();
  p[0] = read();
  while (p[cnt] != 1) p[cnt + 1] = phi(p[cnt]), cnt++;
  for (int i = 1; i <= n; i++) a[i] = read();
  int q = read();
  for (int i = 1; i <= q; i++) {
    int l = read(), r = read();
    printf("%lld\n", Mi(l, r, 0) % p[0]);
  }
  return 0;
}