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 N = 500005, M = 20000005;
long long Q, n;
long long a[N], P;
map<long long, long long> mp;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
inline long long FUCK(long long x, long long p) {
  return x < p ? x : (x % p + p);
}
inline long long ksm(long long x, long long q, long long P) {
  long long r = 1;
  for (; q; q >>= 1, x = FUCK(x * x, P))
    if (q & 1) r = FUCK(r * x, P);
  return r;
}
inline long long William(long long x) {
  long long r = x, xx = x;
  if (mp.count(x)) return mp[x];
  for (long long i = 2; i <= sqrt(x); i++)
    if (!(x % i)) {
      r = r / i * (i - 1);
      while (!(x % i)) x /= i;
    }
  if (x > 1)
    return mp[xx] = r / x * (x - 1);
  else
    return mp[xx] = r;
}
inline long long Nephren(long long l, long long r, long long P) {
  if (l == r || P == 1)
    return FUCK(a[l], P);
  else
    return ksm(a[l], Nephren(l + 1, r, William(P)), P);
}
signed main() {
  read(n), read(P);
  for (long long i = 1; i <= n; i++) read(a[i]);
  for (read(Q); Q--;) {
    long long l, r;
    read(l), read(r);
    printf("%lld\n", (Nephren(l, r, P) % P + P) % P);
  }
  return 0;
}