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 = 1200005;
int n, q;
vector<int> v[N];
int f[N * 5][8], x, ans[15], Ans[N];
struct Ask {
  int l, r, from;
} A[N];
inline int cmp(Ask x, Ask y) { return x.r < y.r; }
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++)
      if (x % j == 0) {
        int flag = 0;
        while (x % j == 0) flag ^= 1, x /= j;
        if (flag) v[i].push_back(j);
      }
    if (x != 1) v[i].push_back(x);
  }
  for (int i = 1; i <= q; i++) scanf("%d%d", &A[i].l, &A[i].r), A[i].from = i;
  sort(A + 1, A + q + 1, cmp);
  for (int i = 1, l = 1; i <= n; i++) {
    for (int j = 0; j < 1 << v[i].size(); j++) {
      int cnt = 0, s = 1;
      for (int k = 0; k < v[i].size(); k++)
        if ((1 << k) & j)
          s *= v[i][k];
        else
          cnt++;
      for (int k = 0; k <= 7; k++) ans[k + cnt] = max(ans[k + cnt], f[s][k]);
      f[s][cnt] = i;
    }
    while (l <= q && i == A[l].r) {
      for (int j = 0; j <= 14; j++)
        if (ans[j] >= A[l].l) {
          Ans[A[l].from] = j;
          break;
        }
      l++;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", Ans[i]);
}