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>
namespace ringo {
template <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  register bool f = 0;
  while (!isdigit(c)) f ^= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  if (f) x = -x;
}
template <class T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar('0' + x % 10);
}
template <class T>
inline void print(T x, char c) {
  print(x), putchar(c);
}
const int N = 5e5 + 10, L = 60;
int n, m, lim, end, rit, lst, c_pri, a[N], mod[L], ans[L];
bool flag[L];
struct node {
  int l, r, mid, fst;
} p[N << 2];
inline int phi(int x) {
  int ans = 1;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      x /= i, ans *= i - 1;
      while (x % i == 0) x /= i, ans *= i;
    }
  if (x != 1) ans *= x - 1;
  return ans;
}
inline void push(long long x) {
  while (~end)
    if (x >= mod[end + 1])
      flag[end--] = 1;
    else
      break;
}
inline int fpow(int a, int b, int p) {
  int s = 1;
  a %= p;
  for (; b; b >>= 1, a = (long long)a * a % p)
    if (b & 1) s = (long long)s * a % p;
  return s;
}
inline int fpow(int a, int b) {
  int s = 1;
  push(a), a %= mod[0];
  while (b) {
    if (b & 1) push((long long)s * a), s = (long long)s * a % mod[0];
    b >>= 1;
    if (b) push((long long)a * a), a = (long long)a * a % mod[0];
  }
  return s;
}
void build(int l, int r, int u = 1) {
  p[u].l = l, p[u].r = r, p[u].mid = (l + r) >> 1;
  if (l == r) {
    p[u].fst = a[l] == 1 ? l : n + 2;
    return;
  }
  build(l, p[u].mid, u << 1), build(p[u].mid + 1, r, u << 1 | 1);
  p[u].fst = std::min(p[u << 1].fst, p[u << 1 | 1].fst);
}
inline int query(int k) { return a[k]; }
int query_fst(int l, int r, int u = 1) {
  if (p[u].l == l && p[u].r == r) return p[u].fst;
  if (r <= p[u].mid) return query_fst(l, r, u << 1);
  if (l > p[u].mid) return query_fst(l, r, u << 1 | 1);
  return std::min(query_fst(l, p[u].mid, u << 1),
                  query_fst(p[u].mid + 1, r, u << 1 | 1));
}
int solve(int l, int r, int p, int lim) {
  memset(flag, 0, sizeof(flag));
  rit = query_fst(l, r) - 1, lst = 1;
  if (rit == n + 1) rit = r;
  if (rit < l) return query(l) % mod[0];
  r = rit, end = lim;
  while (rit >= l) {
    if (flag[0]) break;
    end = std::min(end, rit - l - 1);
    lst = fpow(query(rit--), lst);
  }
  if (l + lim < r)
    ans[lim] = 1;
  else
    lim = r - l, ans[lim] = query(r);
  ans[lim] %= mod[lim];
  for (int i = lim - 1; i >= 0; i--) {
    ans[i] =
        fpow(query(l + i), ans[i + 1] + (flag[i] ? mod[i + 1] : 0), mod[i]);
  }
  return ans[0] % mod[0];
}
void main() {
  read(n);
  int p;
  read(p);
  for (int i = 1; i <= n; i++) read(a[i]);
  build(1, n);
  mod[lim = 0] = p;
  for (int i = 1; mod[i - 1] != 1; i++) mod[i] = phi(mod[i - 1]), ++lim;
  mod[lim + 1] = 1;
  read(m);
  for (int i = 1, l, r; i <= m; i++) {
    read(l), read(r);
    print(solve(l, r, p, lim), '\n');
  }
}
}  // namespace ringo
signed main() { return ringo::main(), 0; }