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 = 194600, M = 5032110;
int n, q, a[N], m, vis[M], f[N][15], g[15];
vector<int> v[M];
void dfs(int* p, int c, int x, int s) {
  if (c--)
    dfs(p, c, x, s + 1), dfs(p, c, x * p[c], s);
  else {
    for (int i = v[x].size(); i--;) {
      int t = (s & 15) + (v[x][i] & 15);
      if (t < 15 && g[t] < v[x][i] >> 4) g[t] = v[x][i] >> 4;
      if ((v[x][i] & 15) >= (s & 15)) v[x].pop_back();
    }
    v[x].push_back(s);
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), a[i] > m ? m = a[i] : 1;
  for (int i = 2; i <= m; i++)
    if (!vis[i])
      for (int j = i; j <= m; j += i)
        if (!vis[j]) vis[j] = i;
  for (int i = 0; i < 15; i++) g[i] = -1;
  for (int i = 0; i < n; i++) {
    int p[10], c = 0;
    for (int x = a[i]; x > 1;) {
      int y = vis[x], z = 0;
      while (x % y == 0) x /= y, z ^= 1;
      if (z) p[c++] = y;
    }
    dfs(p, c, 1, i << 4);
    for (int j = 0; j < 15; j++) f[i + 1][j] = g[j];
  }
  while (q--) {
    int l, r, s = 0;
    scanf("%d%d", &l, &r);
    --l;
    while (f[r][s] < l) s++;
    printf("%d\n", s);
  }
  return 0;
}