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 = 1e5 + 5;
const int INF = 0x3f3f3f3f;
int N, M, Q;
int w[MAXN];
vector<int> f;
set<int> ones;
int lft[MAXN][10];
void load() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; i++) scanf("%d", w + i);
}
inline int mul(int x, int y, int mod) { return (long long)x * y % mod; }
int pot(int x, int y, int mod) {
  int res = 1;
  for (; y; y /= 2) {
    if (y % 2) res = mul(res, x, mod);
    x = mul(x, x, mod);
  }
  return res;
}
int query(int from, int to, int dep) {
  if (f[dep] == 1) return 0;
  if (from == to) return w[from] % f[dep];
  int nxt = query(from + 1, to, dep + 1);
  return pot(w[from],
             to - from >= 5 || lft[to][to - from] == INF ? nxt + f[dep + 1]
                                                         : lft[to][to - from],
             f[dep]);
}
int phi(int x) {
  int res = x;
  for (int i = 2; i * i <= x; i++) {
    if (x % i) continue;
    while (!(x % i)) x /= i;
    res /= i;
    res *= i - 1;
  }
  if (x > 1) {
    res /= x;
    res *= x - 1;
  }
  return res;
}
int naive(int x, int y) {
  if (y > 30) return INF;
  long long res = 1;
  while (y--) {
    res *= x;
    if (res > INF) return INF;
  }
  return res;
}
void solve() {
  f.push_back(M);
  while (f.back() > 1) f.push_back(phi(f.back()));
  for (int i = 1; i <= N; i++) {
    lft[i][0] = 1;
    for (int j = 1; j <= min(i, 5); j++)
      lft[i][j] = naive(w[i - j + 1], lft[i][j - 1]);
  }
  for (int i = 1; i <= N; i++)
    if (w[i] == 1) ones.insert(i);
  scanf("%d", &Q);
  while (Q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (w[l] == 1) {
      printf("%d\n", 1 % M);
      continue;
    }
    auto it = ones.lower_bound(l);
    if (it != ones.end()) r = min(r, *it - 1);
    printf("%d\n", query(l, r, 0));
  }
}
int main() {
  load();
  solve();
  return 0;
}