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;
int l[5032107 + 5][8], ans[200009][20];
vector<int> primes;
bool isP[5032107 + 5];
void crivo() {
  for (int i = 2; i < 5032107 + 5; i++) isP[i] = true;
  for (int i = 2; i < 5032107 + 5; i++) {
    if (!isP[i]) continue;
    primes.push_back(i);
    for (int j = i + i; j < 5032107 + 5; j += i) isP[j] = false;
  }
}
int main() {
  int n, q, v;
  scanf("%d %d", &n, &q);
  crivo();
  for (int idx = 1; idx <= n; idx++) {
    scanf("%d", &v);
    vector<int> fac;
    for (int i = 0; i < int(primes.size()) && primes[i] * primes[i] <= v; i++) {
      if (v % primes[i] != 0) continue;
      int cnt = 0;
      while (v % primes[i] == 0) {
        v /= primes[i];
        cnt++;
      }
      if (cnt % 2) fac.push_back(primes[i]);
    }
    if (v > 1) fac.push_back(v);
    int k = int(fac.size());
    for (int j = 0; j < 15; j++) ans[idx][j] = ans[idx - 1][j];
    for (int i = 0; i < (1 << k); i++) {
      int p = 1, cnt = 0;
      for (int j = 0; j < k; j++) {
        if ((i >> j) & 1)
          p *= fac[j];
        else
          cnt++;
      }
      for (int j = 0; j < 8; j++) {
        ans[idx][j + cnt] = max(ans[idx][j + cnt], l[p][j]);
      }
      l[p][cnt] = idx;
    }
  }
  while (q--) {
    int l, r, a;
    scanf("%d %d", &l, &r);
    a = 0;
    while (ans[r][a] < l) a++;
    printf("%d\n", a);
  }
}