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 = 200005;
int n, q;
int best[15][N], dp[9][5100005];
vector<int> pr[N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    for (int j = 2; j * j <= x; ++j) {
      if (x % j == 0) {
        int c = 0;
        while (x % j == 0) x /= j, ++c;
        if (c & 1) pr[i].push_back(j);
      }
    }
    if (x > 1) pr[i].push_back(x);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 15; ++j) best[j][i] = best[j][i - 1];
    int ST = 1 << pr[i].size();
    for (int s = 0; s < ST; ++s) {
      int val = 1, c = pr[i].size();
      for (int j = 0; j < pr[i].size(); ++j) {
        if ((s >> j) & 1) val *= pr[i][j], --c;
      }
      for (int j = 0; j < 8; ++j) {
        best[j + c][i] = max(best[j + c][i], dp[j][val]);
      }
      dp[c][val] = max(dp[c][val], i);
    }
  }
  for (int l, r; q; --q) {
    scanf("%d%d", &l, &r);
    for (int i = 0; i < 15; ++i) {
      if (best[i][r] >= l) {
        printf("%d\n", i);
        break;
      }
    }
  }
  return 0;
}