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;
int n, q, a[200010], f[5100010][10], best[20];
int l[1100010], ans[1100010];
vector<int> p[200010], v[200010];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int t = a[i];
    for (int j = 2; j <= sqrt(t); j++) {
      int s = 0;
      while (t % j == 0) {
        s++;
        t /= j;
      }
      if (s & 1) {
        p[i].push_back(j);
      }
    }
    if (t > 1) {
      p[i].push_back(t);
    }
  }
  for (int i = 1; i <= q; i++) {
    int x;
    scanf("%d%d", &l[i], &x);
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    int sz = p[i].size();
    for (int mask = 0; mask < (1 << sz); mask++) {
      int x = 1, y = 0;
      for (int j = 0; j < sz; j++) {
        if (mask & (1 << j)) {
          x *= p[i][j];
        } else {
          y++;
        }
      }
      for (int j = 0; j < 8 && j + y < 12; j++) {
        best[j + y] = max(best[j + y], f[x][j]);
      }
    }
    for (int j = 0; j < v[i].size(); j++) {
      int k = 0;
      while (best[k] < l[v[i][j]]) {
        k++;
      }
      ans[v[i][j]] = k;
    }
    for (int mask = 0; mask < (1 << sz); mask++) {
      int x = 1, y = 0;
      for (int j = 0; j < sz; j++) {
        if (mask & (1 << j)) {
          x *= p[i][j];
        } else {
          y++;
        }
      }
      f[x][y] = i;
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}