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
#include <bits/stdc++.h>
using namespace std;
template <typename T = long long>
inline T read() {
  T s = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 3) + (s << 1) + ch - 48;
    ch = getchar();
  }
  return s * f;
}
const int N = 1e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7, CM = 998244353,
          INF = 0x3f3f3f3f;
std::map<int, int> mp;
int phi(int n) {
  if (mp.count(n)) return mp[n];
  int t = 1, i;
  if (!(n & 1))
    for (n >>= 1; !(n & 1); n >>= 1, t <<= 1)
      ;
  for (int i = 3; i * i <= n; i++)
    if (n % i == 0)
      for (n /= i, t *= i - 1; n % i == 0; n /= i, t *= i)
        ;
  if (n > 1) t *= n - 1;
  return mp[n] = t;
}
pair<long long, bool> ksm(long long a, long long b, long long p) {
  pair<long long, bool> ans = {1, 0};
  while (b) {
    if (b & 1) {
      if (ans.first * a >= p) ans.second = 1;
      ans.first = ans.first * a % p;
    }
    if (a * a >= p) ans.second = 1;
    a = a * a % p;
    b >>= 1;
  }
  return ans;
}
int a[N];
pair<long long, bool> cal(long long pos, long long b, long long p) {
  if (p == 1) return {1, 1};
  if (pos == b + 1) return {1, 0};
  long long ph = phi(p);
  pair<long long, bool> ss = cal(pos + 1, b, ph);
  return ksm(a[pos], ss.second ? ss.first % ph + ph : ss.first % ph, p);
}
void solve(int kase) {
  int n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  int q = read();
  for (int i = 1; i <= q; i++) {
    int l = read(), r = read();
    printf("%lld\n", cal(l, r, m).first % m);
  }
}
const bool DUO = 0;
int main() {
  clock_t start, finish;
  double totaltime;
  start = clock();
  if (DUO) {
    int Kase = 0;
    cin >> Kase;
    for (int kase = 1; kase <= Kase; kase++) solve(kase);
  } else
    solve(1);
  finish = clock();
  return 0;
  printf("\nTime: %lfms\n", (double)(finish - start) / CLOCKS_PER_SEC * 1000);
  return 0;
}
