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 = 195000, Q = 1050000;
int n, q, a[N], p[N][8];
int l[Q], t[N], ss[20];
vector<int> no[N];
int dp[Q * 5][8], ans[Q];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    int x, x1, y;
    scanf("%d", &x);
    x1 = x;
    for (int j = 2; j * j <= x1; j++) {
      for (y = 0; x % j == 0; x /= j, y ^= 1)
        ;
      if (y) p[i][t[i]++] = j;
    }
    if (x > 1) p[i][t[i]++] = x;
  }
  for (int i = 1, x; i <= q; i++) scanf("%d%d", &l[i], &x), no[x].push_back(i);
  for (int i = 1; i <= n; i++) {
    for (int s = 0; s < 1 << t[i]; s++) {
      int x = 1, x1 = 0;
      for (int j = 0; j < t[i]; j++)
        if ((s >> j) & 1)
          x *= p[i][j];
        else
          x1++;
      for (int j = 0; j < 8 && j + x1 < 12; j++)
        ss[j + x1] = max(ss[j + x1], dp[x][j]);
    }
    for (auto j : no[i])
      for (int k = 0;; k++)
        if (ss[k] >= l[j]) {
          ans[j] = k;
          break;
        }
    for (int s = 0; s < 1 << t[i]; s++) {
      int x = 1, x1 = 0;
      for (int j = 0; j < t[i]; j++)
        if ((s >> j) & 1)
          x *= p[i][j];
        else
          x1++;
      dp[x][x1] = i;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}