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 fk = 5032108;
int n, q, a[200007], l, r, prime[fk];
int ret[fk][8], dp[200007][15];
vector<int> pr[200007];
vector<int> prep(int x) {
  int cur = 0, cnt = 0;
  vector<int> res;
  while (x > 1) {
    if (prime[x] != cur) {
      if (cnt % 2 == 1) res.push_back(cur);
      cur = prime[x];
      cnt = 0;
    }
    cnt++;
    x /= prime[x];
  }
  if (cnt % 2 == 1) res.push_back(cur);
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 2; i < fk; i++) {
    if (prime[i]) continue;
    for (int j = i; j < fk; j += i) prime[j] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pr[i] = prep(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 14; j++) dp[i][j] = dp[i - 1][j];
    int sz = (int)pr[i].size();
    for (int mask = 0; mask < (1 << sz); mask++) {
      int add = 0, prd = 1;
      for (int j = 0; j < sz; j++) {
        if ((mask & (1 << j)))
          prd *= pr[i][j];
        else
          add++;
      }
      for (int j = 0; j <= 7; j++) {
        dp[i][add + j] = max(dp[i][add + j], ret[prd][j]);
      }
      ret[prd][add] = i;
    }
  }
  for (int t = 1; t <= q; t++) {
    cin >> l >> r;
    int ans = 0;
    for (int i = 0; i <= 14; i++) {
      if (dp[r][i] >= l) {
        ans = i;
        break;
      }
    }
    cout << ans << "\n";
  }
}