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 inf = 1012345678;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int N, Q;
  cin >> N >> Q;
  vector<vector<int> > g(N);
  int limit = 0;
  for (int i = 0; i < N; ++i) {
    int x, rad = 1;
    cin >> x;
    for (int j = 2; j * j <= x; ++j) {
      int cnt = 0;
      while (x % j == 0) x /= j, cnt ^= 1;
      if (cnt == 1) g[i].push_back(j), rad *= j;
    }
    if (x > 1) g[i].push_back(x), rad *= x;
    limit = max(limit, rad);
  }
  vector<vector<pair<int, int> > > qs(N + 1);
  for (int i = 0; i < Q; ++i) {
    int L, R;
    cin >> L >> R;
    --L;
    qs[R].push_back(make_pair(L, i));
  }
  vector<vector<int> > curopt(limit + 1, vector<int>(8, -inf));
  vector<int> curbound(15, -inf);
  vector<int> ans(Q);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < 1 << g[i].size(); ++j) {
      int rem = 0, mul = 1;
      for (int k = 0; k < g[i].size(); ++k) {
        if ((j >> k) & 1)
          mul *= g[i][k];
        else
          ++rem;
      }
      for (int k = 0; k <= 7; ++k) {
        curbound[rem + k] = max(curbound[rem + k], curopt[mul][k]);
      }
      curopt[mul][rem] = i;
    }
    for (pair<int, int> j : qs[i + 1]) {
      for (int k = 0; k <= 14; ++k) {
        if (curbound[k] >= j.first) {
          ans[j.second] = k;
          break;
        }
      }
    }
  }
  for (int i = 0; i < Q; ++i) {
    cout << ans[i] << '\n';
  }
  return 0;
}