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 mod = (1e9) + 7;
const double eps = 1e-6;
const int siz = 194599, siz2 = 1049659, siz3 = 5032108;
int n, q, arr[siz], lft[siz2], dp[siz3][8], best[15], ans[siz2];
vector<int> inds[siz], primes;
void factorize(int v) {
  for (int i = 2; i * i <= v; i++) {
    if (v % i == 0) {
      int c = 0;
      while (v % i == 0) {
        v /= i;
        c++;
      }
      if (c % 2) {
        primes.push_back(i);
      }
    }
  }
  if (v > 1) {
    primes.push_back(v);
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    inds[r].push_back(i);
    lft[i] = l;
  }
  for (int i = 1; i <= n; i++) {
    factorize(arr[i]);
    int sz = primes.size();
    int lim = 1 << sz;
    for (int j = 0; j < lim; j++) {
      int mask = j, setbits = 0, val = 1;
      for (int k = 0;; k++) {
        if (mask == 0) {
          break;
        }
        int mask2 = 1 << k;
        if (mask & mask2) {
          val *= primes[k];
          mask &= ~mask2;
          setbits++;
        }
      }
      int unset = sz - setbits;
      for (int k = 0; k <= 7; k++) {
        best[k + unset] = max(best[k + unset], dp[val][k]);
      }
    }
    for (auto ind : inds[i]) {
      for (int j = 0; j <= 14; j++) {
        if (best[j] >= lft[ind]) {
          ans[ind] = j;
          break;
        }
      }
    }
    for (int j = 0; j < lim; j++) {
      int mask = j, setbits = 0, val = 1;
      for (int k = 0;; k++) {
        if (mask == 0) {
          break;
        }
        int mask2 = 1 << k;
        if (mask & mask2) {
          val *= primes[k];
          mask &= ~mask2;
          setbits++;
        }
      }
      int unset = sz - setbits;
      dp[val][unset] = i;
    }
    primes.clear();
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}