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>
char buf[1 << 21], *p1 = buf, *p2 = buf;
using namespace std;
long long n, m, Mod, a[100000 + 5];
map<long long, long long> phi;
template <typename T>
void Read(T &x) {
  x = 0;
  long long fu = 1;
  char ch =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
           ? EOF
           : *p1++);
  for (; !isdigit(ch);
       ch = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                 ? EOF
                 : *p1++))
    if (ch == '-') fu = -1;
  for (; isdigit(ch);
       ch = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                 ? EOF
                 : *p1++))
    x = (x << 3) + (x << 1) + (ch - 48);
  x *= fu;
}
long long Phi(long long x) {
  long long res = x, pos = x;
  if (phi.count(x)) return phi[x];
  for (long long i = 2; i <= sqrt(x); i++) {
    if (!(x % i)) {
      res = res / i * (i - 1);
      while (!(x % i)) x /= i;
    }
  }
  if (x > 1) res = res / x * (x - 1);
  return phi[pos] = res;
}
long long Mo(long long x, long long p) { return x < p ? x : (x % p + p); }
long long QuickPow(long long x, long long y, long long p) {
  long long res = 1;
  while (y) {
    if (y & 1) res = Mo(res * x, p);
    x = Mo(x * x, p);
    y >>= 1;
  }
  return res;
}
long long Dfs(long long l, long long r, long long p) {
  return (l == r || p == 1) ? Mo(a[l], p)
                            : QuickPow(a[l], Dfs(l + 1, r, Phi(p)), p);
}
signed main() {
  Read(n);
  Read(Mod);
  for (long long i = 1; i <= n; i++) Read(a[i]);
  Read(m);
  while (m--) {
    long long x, y;
    Read(x);
    Read(y);
    printf("%lld\n", Dfs(x, y, Mod) % Mod);
  }
}