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>
#pragma GCC optimize(3)
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
  Finish_read = 0;
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    if (ch == EOF) return;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
  Finish_read = 1;
}
template <class T>
inline void print(T x) {
  if (x / 10 != 0) print(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void writeln(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
  putchar('\n');
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
}
const int maxn = 100010;
vector<long long> phi;
int n, m, prime[maxn], cnt;
long long a[maxn], mod;
bool vis[maxn];
inline void init() {
  for (int i = 2; i <= 100000; i++) {
    if (!vis[i]) prime[++cnt] = i;
    for (int j = 1; i * prime[j] <= 100000 && j <= cnt; j++) {
      int t = i * prime[j];
      vis[t] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
inline long long pw(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
inline long long getphi(long long n) {
  long long ans = n;
  for (int i = 1; i <= cnt; ++i) {
    if (n % prime[i] == 0) {
      ans -= ans / prime[i];
      while (n % prime[i] == 0) n /= prime[i];
    }
  }
  if (n > 1) ans -= ans / n;
  return ans;
}
inline long long solve(int l, int r, int k) {
  if (l == r) return min(a[l], a[l] % phi[k] + phi[k]);
  if (phi[k] == 1) return 1;
  long long o = solve(l + 1, r, k + 1);
  if (a[l] == 1) return 1;
  long long v = pw(a[l], o, phi[k]), d = 1ll << 60, e = 1;
  for (int i = 0; i < o; i++) {
    d /= a[l];
    e *= a[l];
    if (d == 0) break;
  }
  if (d == 0)
    return v + phi[k];
  else
    return min(e, e % phi[k] + phi[k]);
}
int main() {
  init();
  read(n);
  read(mod);
  phi.push_back(mod);
  while (phi[phi.size() - 1] != 1) phi.push_back(getphi(phi[phi.size() - 1]));
  for (int i = 1; i <= n; i++) read(a[i]);
  read(m);
  while (m--) {
    int l, r;
    read(l);
    read(r);
    printf("%lld\n", solve(l, r, 0) % phi[0]);
  }
}