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;
map<int, int> memo;
vector<long long> w;
int phiF(int n) {
  auto a = memo.find(n);
  if (a != memo.end()) return a->second;
  int nn = n;
  if (n <= 1) {
    memo[n] = n;
    return n;
  }
  int ans = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      ans -= ans / i;
    }
  }
  if (n > 1) ans -= ans / n;
  memo[nn] = ans;
  return ans;
}
long long expMod(long long a, long long exp, long long mod) {
  if (exp == 0) return 1;
  long long pa = expMod(a, exp / 2, mod);
  pa = (pa * pa) % mod;
  if (exp % 2 == 1) pa = (pa * a) % mod;
  return pa;
}
bool isBig(int l, int r, long long lim, long long &res) {
  if (w[l] == 1) {
    res = 1;
    return false;
  }
  if (l == r) {
    res = w[l];
    return w[l] >= lim;
  }
  if (w[l] >= lim) return true;
  long long wp = 1;
  int e = 0;
  while (wp < lim) {
    wp *= w[l];
    e++;
  }
  long long rr;
  if (isBig(l + 1, r, e, rr)) return true;
  res = 1;
  for (int i = 0; i < rr; i++) res *= w[l];
  return false;
}
long long solve(int l, int r, long long mod) {
  if (l == r) return w[l] % mod;
  if (mod == 1) return 0;
  long long res;
  long long lim = 0, mm = mod;
  while (mm) {
    mm >>= 1;
    lim++;
  }
  if (isBig(l + 1, r, lim, res)) {
    long long m2 = phiF(mod);
    long long e = solve(l + 1, r, m2) + m2;
    return expMod(w[l], e, mod);
  } else {
    return expMod(w[l], res, mod);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  w.resize(n);
  for (int i = 0; i < n; i++) cin >> w[i];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    auto ans = solve(l, r, m);
    cout << ans << endl;
  }
  return 0;
}