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 int MAXN = 1e6 + 10;
long long Fc[MAXN], Prime[MAXN], CntPrime, Phi[MAXN];
vector<int> v;
inline void sieve() {
  Phi[1] = 1;
  for (int i = 2; i <= 1e6; i++) {
    if (!Fc[i]) Phi[i] = (Prime[++CntPrime] = Fc[i] = i) - 1;
    for (int j = 1; j <= CntPrime; j++)
      if (Prime[j] > Fc[i] || Prime[j] * i > 1e6)
        break;
      else
        Fc[Prime[j] * i] = Prime[j],
                      Phi[Prime[j] * i] =
                          Phi[i] * (Prime[j] - (Fc[i] != Prime[j]));
  }
}
inline int phi(int num) {
  if (num <= 1e6) return Phi[num];
  int Ans = num, Val = num;
  for (int i = 1; Prime[i] * Prime[i] <= num; i++)
    if (num % Prime[i] == 0) {
      Ans = Ans / Prime[i] * (Prime[i] - 1);
      while (Val % Prime[i] == 0) Val /= Prime[i];
    }
  if (Val ^ 1) Ans = Ans / Val * (Val - 1);
  return Ans;
}
int N, M, W[MAXN];
inline long long fpow(long long a, long long x, long long mod) {
  long long ans = 1;
  bool geq = (a >= mod && x >= 1);
  for (a %= mod; x; x >>= 1, geq |= (a * a >= mod && x >= 1), a = a * a % mod)
    if (x & 1) {
      geq |= (ans * a >= mod);
      ans = ans * a % mod;
    }
  return ans + geq * mod;
}
inline long long ans(int l, int r) {
  r = min(l + (int)v.size() - 1, r);
  long long val = W[r] % v[r - l] + (W[r] >= v[r - l]) * v[r - l];
  for (int i = r - l - 1; i >= 0; i--) val = fpow(W[i + l], val, v[i]);
  return val % v[0];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve();
  cin >> N >> M;
  for (; M ^ 1; M = phi(M)) v.push_back(M);
  v.push_back(1);
  for (int i = 1; i <= N; i++) cin >> W[i];
  int Q;
  cin >> Q;
  for (int i = 1, l, r; i <= Q; i++) {
    cin >> l >> r;
    cout << ans(l, r) << "\n";
  }
  cout.flush();
  return 0;
}