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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
const size_t MAX = 2e5 + 5;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int Move[][3] = {{0, 0}, {1, 0},  {0, 1},  {0, -1}, {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
inline long long gcd(const long long& a, const long long& b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long lcm(const long long& a, const long long& b) {
  return a / gcd(a, b) * b;
}
inline long long sqr(const long long& x) { return x * x; }
long long Pow(long long a, int t) {
  a %= MOD;
  long long base = 1;
  while (t > 0) {
    if (t & 1) base = base * a % MOD;
    a = a * a % MOD;
    t >>= 1;
  }
  return base;
}
template <class T>
inline bool chkmax(T& a, T b) {
  return (a < b) ? (a = b, 1) : 0;
}
template <class T>
inline bool chkmin(T& a, T b) {
  return (a > b) ? (a = b, 1) : 0;
}
long long eular(long long x) {
  long long ret = x, a = x;
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      ret = ret / i * (i - 1);
      while (a % i == 0) a /= i;
    }
  }
  return (a > 1) ? (ret / a * (a - 1)) : ret;
}
inline long long Mod(long long a, long long mod) {
  return a < mod ? a : (a % mod + mod);
}
long long qpow(long long a, long long t, long long mod) {
  long long base = 1;
  while (t > 0) {
    if (t & 1) base = Mod(base * a, mod);
    a = Mod(a * a, mod);
    t >>= 1;
  }
  return base;
}
long long phi[MAX], tot;
long long v[MAX];
long long solve(int l, int r, int i) {
  if (l == r || phi[i] == 1) return Mod(v[l], phi[i]);
  return qpow(v[l], solve(l + 1, r, i + 1), phi[i]);
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (phi[0] = m, tot = 0; phi[tot] != 1; tot++) {
    phi[tot + 1] = eular(phi[tot]);
  }
  for (int i = 1; i <= n; i++) cin >> v[i];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r, 0) % m << '\n';
  }
}
