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;
const int MAX = 5032110;
int n, q;
vector<int> pr[N];
int prv[MAX][8], f[N][15];
vector<int> primes(int a) {
  vector<int> ret;
  for (int i = 2; i * i <= a; ++i)
    if (a % i == 0) {
      while (a % (i * i) == 0) a /= (i * i);
      if (a % i == 0) a /= i, ret.push_back(i);
    }
  if (a > 1) ret.push_back(a);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    int a;
    cin >> a;
    pr[i] = primes(a);
  }
  for (int i = 1; i <= n; ++i) {
    if (i > 1)
      for (int j = 0; j < 15; ++j) f[i][j] = f[i - 1][j];
    int sz = pr[i].size();
    for (int u = 0; u < (1 << sz); ++u) {
      int prd = 1;
      for (int v = 0; v < sz; ++v)
        if (u & (1 << v)) {
          prd *= pr[i][v];
        }
      int d = sz - __builtin_popcount(u);
      for (int dd = 0; dd <= 7; ++dd) {
        f[i][d + dd] = max(f[i][d + dd], prv[prd][dd]);
      }
      prv[prd][d] = i;
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    for (int i = 0; i < 15; ++i)
      if (f[r][i] >= l) {
        cout << i << '\n';
        break;
      }
  }
}