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 long long M = 1e5 + 3;
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline long long read() {
  long long x = 0, f = 1;
  char c =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
           ? EOF
           : *p1++);
  for (; !isdigit(c);
       c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    if (c == '-') f = -1;
  for (; isdigit(c);
       c = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    x = x * 10 + c - '0';
  return x * f;
}
char sr[1 << 21], z[20];
long long C = -1, Z;
inline void Ot() { fwrite(sr, 1, C + 1, stdout), C = -1; }
inline void print(long long x, char chr = '\n') {
  if (C > 1 << 20) Ot();
  if (x < 0) sr[++C] = 45, x = -x;
  while (z[++Z] = x % 10 + 48, x /= 10)
    ;
  while (sr[++C] = z[Z], --Z)
    ;
  sr[++C] = chr;
}
long long n, mod, a[M], phi[35];
inline long long Phi(long long x) {
  long long res = x, pos = x;
  for (long long i = 2; i <= sqrt(x); ++i)
    if (!(x % i))
      for (res = res / i * (i - 1); !(x % i); x /= i)
        ;
  if (x > 1) res = res / x * (x - 1);
  return res;
}
inline long long Mo(long long x, long long p) {
  return x < p ? x : (x % p + p);
}
inline long long qpow(long long x, long long p, long long mod) {
  long long s = 1;
  for (; p; p >>= 1, x = Mo(x * x, mod))
    if (p & 1) s = Mo(s * x, mod);
  return s;
}
long long dfs(long long l, long long r, long long i) {
  return (l == r || phi[i] == 1) ? Mo(a[l], phi[i])
                                 : qpow(a[l], dfs(l + 1, r, i + 1), phi[i]);
}
signed main() {
  n = read(), mod = read(), phi[0] = mod;
  for (long long i = 1; i <= 30; ++i) phi[i] = Phi(phi[i - 1]);
  for (long long i = 1; i <= n; ++i) a[i] = read();
  for (long long m = read(), l, r; m; --m)
    l = read(), r = read(), print(dfs(l, r, 0) % mod);
  return Ot(), 0;
}