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;
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
const long long N = 500005, p = 2000000009;
long long r, k, n, c[N], m, a[N];
map<long long, long long> phi;
inline long long ksm(long long a, long long b, long long mod) {
  a %= mod;
  long long t = 1, y = a;
  while (b) {
    if (b & 1) t = (long long)t * y % mod;
    y = (long long)y * y % mod;
    b >>= 1;
  }
  return t;
}
inline long long check(long long l, long long r) {
  if (l > r) return 1;
  if (l == r) return a[l];
  long long zs = a[r];
  for (int i = l; i <= r; i++)
    if (a[i] == 1) return check(l, i - 1);
  for (long long i = r - 1; i >= l; i--)
    if (a[i] > 1) {
      if (log(p) / log(a[i]) < zs - 1)
        return p + 1;
      else
        zs = ksm(a[i], zs, 1e18);
    } else
      zs = 1;
  return zs;
}
long long solve(long long l, long long r, long long mod) {
  if (a[l] % mod == 0)
    return 0;
  else if (l + 1 == r)
    return ksm(a[l], a[r], mod);
  else if (l == r)
    return a[l] % mod;
  long long t = check(l + 1, min(l + 5, r));
  if (t > phi[mod])
    return ksm(a[l], solve(l + 1, r, phi[mod]) % phi[mod] + phi[mod], mod);
  else
    return ksm(a[l], t, mod);
}
inline long long calc(long long x) {
  long long zs = x, ans = x;
  for (long long i = 2; i * i <= x; i++)
    if (zs % i == 0) {
      ans = ans / i * (i - 1);
      while (zs % i == 0) zs /= i;
    }
  if (zs > 1) ans = ans / zs * (zs - 1);
  return ans;
}
int main() {
  n = read();
  long long mod = read();
  for (long long i = mod; i > 1; i = phi[i]) phi[i] = calc(i);
  phi[1] = 1;
  for (long long i = 1; i <= n; i++) a[i] = read();
  m = read();
  for (long long i = 1; i <= m; i++) {
    long long s2 = read(), s3 = read();
    long long ans = solve(s2, s3, mod) % mod;
    writeln(ans);
  }
}