Problem: We call an array b_1, b_2, …, b_m good, if there exist two indices i < j such that b_i ⋅ b_j is a [perfect square](https://en.wikipedia.org/wiki/Square_number).

Given an array b_1, b_2, …, b_m, in one action you can perform one of the following: 

  * multiply any element b_i by any prime p; 
  * divide any element b_i by prime p, if b_i is divisible by p. 



Let f(b_1, b_2, …, b_m) be the minimum number of actions needed to make the array b good.

You are given an array of n integers a_1, a_2, …, a_n and q queries of the form l_i, r_i. For each query output f(a_{l_i}, a_{l_i + 1}, …, a_{r_i}).

Input

The first line contains two integers n and q (2 ≤ n ≤ 194 598, 1 ≤ q ≤ 1 049 658) — the length of the array and the number of queries.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 5 032 107) — the elements of the array.

Each of the next q lines contains two integers l_i and r_i (1 ≤ l_i < r_i ≤ n) — the parameters of a query.

Output

Output q lines — the answers for each query in the order they are given in the input.

Example

Input

10 10
34 37 28 16 44 36 43 50 22 13
1 3
4 8
6 10
9 10
3 10
8 9
5 6
1 4
1 7
2 6


Output

2
0
1
3
0
1
1
1
0
0

Note

In the first query of the first sample you can multiply second number by 7 to get 259 and multiply the third one by 37 to get 1036. Then a_2 ⋅ a_3 = 268 324 = 518^2.

In the second query subarray is already good because a_4 ⋅ a_6 = 24^2.

In the third query you can divide 50 by 2 to get 25. Then a_6 ⋅ a_8 = 30^2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int M = 5.1e6 + 5;
const int B = 10;
int n, q;
int h[M];
int go[B + 1][N];
vector<int> dv[N];
int best[8][M];
int main() {
  for (int i = 2; i * i < M; i++) {
    if (!h[i]) {
      for (int j = i; j < M; j += i) h[j] = i;
    }
  }
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    while (h[x]) {
      int y = h[x], c = 0;
      while (x % y == 0) {
        x /= y;
        c ^= 1;
      }
      if (c) dv[i].push_back(y);
    }
    if (x > 1) dv[i].push_back(x);
    assert(dv[i].size() <= 7);
  }
  for (int need = 0; need <= B; need++) {
    go[need][n + 1] = n + 1;
    memset(best, 0, sizeof(best));
    for (int i = n; i >= 1; i--) {
      go[need][i] = go[need][i + 1];
      for (int mask = 0; mask < (1 << dv[i].size()); mask++) {
        int prod = 1, cnt = dv[i].size();
        for (int j = 0; j < dv[i].size(); j++) {
          if (mask & (1 << j)) {
            prod *= dv[i][j];
            cnt--;
          }
        }
        if (cnt <= need and need - cnt <= 7 and best[need - cnt][prod]) {
          go[need][i] = min(go[need][i], best[need - cnt][prod]);
        }
        best[cnt][prod] = i;
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    for (int i = 0; i <= B + 1; i++) {
      if (i > B or go[i][l] <= r) {
        printf("%d\n", i);
        break;
      }
    }
  }
  return 0;
}