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;
namespace IO {
int len = 0;
char ibuf[(1 << 20) + 1], *iS, *iT, out[(1 << 27) + 1];
inline int read() {
  register char ch =
      (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
       (iS == iT ? EOF : *iS++) : *iS++);
  register int x = 0;
  while (ch < '0' || ch > '9')
    ch = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
          (iS == iT ? EOF : *iS++) : *iS++);
  while (ch >= '0' && ch <= '9')
    x = x * 10 + (ch ^ 48),
    ch = (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin),
          (iS == iT ? EOF : *iS++) : *iS++);
  return x;
}
inline void write(long long x) {
  if (x > 9) write(x / 10);
  out[len++] = x % 10 + 48;
}
}  // namespace IO
using namespace IO;
unordered_map<int, int> Phi;
struct number {
  long long v;
  bool m;
  number(long long V = 0, bool M = 0) : v(V), m(M) {}
  inline long long f(int mod) { return m ? v + mod : v; }
  inline friend number operator*(const number &a, const number &b) {
    number c;
    c.v = a.v * b.v;
    c.m = a.m | b.m;
    return c;
  }
  inline friend number operator%(const number &a, const int &b) {
    number c = a;
    if (c.v >= b) {
      c.v %= b;
      c.m = 1;
    }
    return c;
  }
};
long long t[100010];
int n, m, mx, mxx, s;
inline int phi(int n) {
  int res, a;
  res = a = n;
  for (int i = 2; i <= sqrt(a); ++i) {
    if (a % i) continue;
    res = res / i * (i - 1);
    while (!(a % i)) a /= i;
  }
  if (a > 1) res = res / a * (a - 1);
  return res;
}
inline void upd(int x, int k) {
  for (int i = x; i <= n; i += (i & -i)) {
    t[i] += k;
  }
}
inline long long query(int x) {
  long long ans = 0;
  for (int i = x; i >= 1; i -= (i & -i)) {
    ans += t[i];
  }
  return ans;
}
inline number qpow(number x, long long y, int mod) {
  number res(1);
  x = x % mod;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
inline number ask(int l, int r, int mod) {
  long long q = query(l);
  if (q == 1) return number(1);
  if (mod == 1) return number(0, 1);
  if (l == r) return number(q % mod, q >= mod);
  return qpow(number(q), ask(l + 1, r, Phi[mod]).f(Phi[mod]), mod);
}
int main() {
  n = read(), mxx = read(), mx = mxx;
  while (mx != 1) {
    Phi[mx] = phi(mx);
    mx = Phi[mx];
  }
  for (int i = 1; i <= n; i++) {
    s = read();
    upd(i, s);
    upd(i + 1, -s);
  }
  m = read();
  for (int i = 1; i <= m; i++) {
    int l = read(), r = read();
    write(ask(l, r, mxx).v % mxx);
    out[len++] = '\n';
  }
  fwrite(out, 1, len, stdout);
  return 0;
}