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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
long long pw(long long n, long long k, long long MOD) {
  long long r(1);
  while (k) {
    if (k & 1) r = r * n >= MOD ? r * n % MOD + MOD : r * n;
    n = n * n >= MOD ? n * n % MOD + MOD : n * n;
    k >>= 1;
  }
  return r;
}
long long P[(1 << 17) >> 1] = {2}, L(1), cn[1 + ((1 << 17) >> 6)];
void gen() {
  (cn[0 >> 6] |= 1ULL << ((0 & 63))), (cn[1 >> 6] |= 1ULL << ((1 & 63)));
  int Q(1 + sqrt((1 << 17)));
  for (long long i(4); i < (1 << 17); i += 2)
    (cn[i >> 6] |= 1ULL << ((i & 63)));
  for (int k(3); k <= Q; k += 2)
    if (!((cn[k >> 6] >> (k & 63)) & 1))
      for (long long h(k << 1), j(k * k); j < (1 << 17); j += h)
        (cn[j >> 6] |= 1ULL << ((j & 63)));
  for (int i(3); i < (1 << 17); i += 2)
    if (!((cn[i >> 6] >> (i & 63)) & 1)) P[L++] = i;
}
int fc(int N, int *f, int *c) {
  int L(0);
  for (int i(0), h(0); N >= P[i] * P[i]; ++i, h = 0) {
    while (!(N % P[i])) ++h, N /= P[i];
    if (h) f[L] = P[i], c[L++] = h;
  }
  if (N > 1) f[L] = N, c[L++] = 1;
  return L;
}
int phi(int N) {
  static int f[64], c[64];
  int L(fc(N, f, c)), P(1);
  for (int i(0); i < L; i++)
    P *= pow(f[i], c[i]) - pow(f[i], c[i] - 1) + (1e-10);
  return P;
}
int N, M, G[(1 << 17)], I, A[(1 << 17)], a, b, S;
int main(void) {
  gen(), scanf("%d%d", &N, &M), *G = M, ga(N, A);
  while (G[I] ^ 1) ++I, G[I] = phi(G[I - 1]);
  int _;
  scanf("%d", &_);
  for (int i(0); i < _; i++) {
    scanf("%d%d", &a, &b), --a, --b, S = 1;
    for (int i = min(b - a, I); ~i; --i) S = pw(A[a + i], S, G[i]);
    printf("%d\n", S % M);
  }
  return 0;
}