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>
int in() {
  int r = 0, c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') return -in();
  for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
    ;
  return r;
}
using namespace std;
vector<int> pri;
vector<long long> pric;
bitset<100000> bi;
int l, r;
int a[100030];
void criba() {
  bi.set();
  for (int i = 2; i < 100000; i++)
    if (bi[i]) {
      for (int j = i + i; j < 100000; j += i) bi[j] = 0;
      pri.push_back(i);
      pric.push_back((long long)i * (long long)i);
    }
}
long long phi(int n) {
  int second = pri.size(), i = 0;
  long long res = n;
  while (pric[i] <= n) {
    if (n % pri[i] == 0) res -= res / pri[i];
    while (n % pri[i] == 0) n /= pri[i];
    i++;
    if (i == second) break;
  }
  if (n > 1) res -= res / n;
  return res;
}
long long modpow(long long base, long long exp, long long modu) {
  if (exp == 0) return 1;
  return (modpow((base * base) % modu, exp / 2, modu) * (exp & 1 ? base : 1)) %
         modu;
}
long long pow(long long x, long long y, long long mod) {
  long long z = 1;
  while (y) {
    if (y & 1) z = z * x;
    if (z > log2(mod)) z = z % mod + mod;
    x = x * x;
    if (x > log2(mod)) x = x % mod + mod;
    y >>= 1;
  }
  return z;
}
long long phipow(long long base, int ind, long long modu) {
  if (modu == 1) return 1;
  if (ind == r) return (base >= modu ? (base % modu + modu) : base);
  long long fi = phi(modu), exp = phipow(a[ind + 1], ind + 1, fi);
  return pow(base, exp, modu);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  criba();
  int n, modu, q;
  cin >> n >> modu;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cin >> q;
  while (q--) {
    cin >> l >> r;
    l--;
    r--;
    cout << phipow(a[l], l, modu) % modu << endl;
  }
  return 0;
}