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 n;
int q;
int msk[200002][20];
int cnt[200002];
vector<pair<int, int> > w[5032109];
vector<pair<int, int> > Q[200002];
int ans[2000002];
int idx[20];
int main() {
  for (int i = 0; i < 2000002; i++) {
    ans[i] = INT_MAX;
  }
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int val;
    scanf("%d", &val);
    for (int z = 2; z * z <= val; z++) {
      bool flag = false;
      while (val % z == 0) {
        flag ^= true;
        val /= z;
      }
      if (flag) {
        msk[i][cnt[i]++] = z;
      }
    }
    if (val > 1) {
      msk[i][cnt[i]++] = val;
    }
  }
  int id = -1;
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    Q[r].push_back(make_pair(l, ++id));
    ans[id] = INT_MAX;
  }
  memset(idx, -1, sizeof(idx));
  for (int i = 0; i < n; i++) {
    sort(Q[i].begin(), Q[i].end());
    for (int j = 0; j < (1 << cnt[i]); j++) {
      int val = 1;
      int er = 0;
      for (int k = 0; k < cnt[i]; k++) {
        if ((j >> k) & 1) {
          val *= msk[i][k];
        } else {
          er++;
        }
      }
      for (auto &lb : w[val]) {
        idx[lb.second + er] = max(idx[lb.second + er], lb.first);
      }
      while (w[val].size() && w[val].back().second >= er) {
        w[val].pop_back();
      }
      w[val].push_back(make_pair(i, er));
    }
    for (auto &g : Q[i]) {
      for (int z = 0; z < 20; z++) {
        if (idx[z] >= g.first) {
          ans[g.second] = z;
          break;
        }
      }
    }
  }
  for (int i = 0; i <= id; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}