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 = 200050;
const int M = 1050000;
const int U = 5 * M;
vector<int> Div[N];
int go[U];
void cut(int x, int id) {
  int c = 0, p = 0;
  while (x > 1) {
    if (go[x] != p) {
      if (c & 1) Div[id].push_back(p);
      c = 0;
    }
    c++;
    p = go[x];
    x /= go[x];
  }
  if (c & 1) Div[id].push_back(p);
}
int sol[20], a[N], ans[M], las[8][U];
vector<pair<int, int> > Qs[N];
int main() {
  int n, i, j, q, l, r, k;
  for (i = 2; i < U; i++) {
    if (!go[i]) {
      for (j = i; j < U; j += i) go[j] = i;
    }
  }
  scanf("%i %i", &n, &q);
  for (i = 1; i <= n; i++) scanf("%i", &a[i]), cut(a[i], i);
  for (i = 1; i <= q; i++)
    scanf("%i %i", &l, &r), Qs[r].push_back(make_pair(l, i));
  for (i = 1; i <= n; i++) {
    int sz = Div[i].size();
    for (j = 0; j < 1 << sz; j++) {
      int b = 1, c = 0;
      for (k = 0; k < sz; k++)
        if ((j >> k) & 1) b *= Div[i][k], c++;
      for (k = 0; k <= 7; k++) {
        sol[k + sz - 2 * c] = max(sol[k + sz - 2 * c], las[k][b]);
      }
      las[sz][b] = i;
    }
    for (j = 0; j < Qs[i].size(); j++) {
      for (k = 19; ~k; k--)
        if (sol[k] >= Qs[i][j].first) ans[Qs[i][j].second] = k;
    }
  }
  for (i = 1; i <= q; i++) printf("%i\n", ans[i]);
  return 0;
}