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 CMAX = int(5e6) + int(1e5);
int n, q, best[15], ans[1100000], a[200000], lst[CMAX][8], rc[CMAX], pk[CMAX];
short cnt[CMAX];
vector<pair<int, int> > qr[200000];
vector<int> p;
void pr(int j, int k, int a, int i) {
  if (j == (int)p.size()) {
    for (int q = 0; q < 8; q++) {
      best[q + cnt[a / k]] = max(best[q + cnt[a / k]], lst[k][q]);
    }
    lst[k][cnt[a / k]] = i;
    return;
  }
  pr(j + 1, k, a, i);
  pr(j + 1, k * p[j], a, i);
}
void relax(int a, int i) {
  p.resize(0);
  a = pk[a];
  int na = a;
  while (a > 1) {
    p.push_back(rc[a]);
    a /= rc[a];
  }
  pr(0, 1, na, i);
}
int main() {
  iostream::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 2; i < CMAX; i++) {
    if (rc[i] == 0) {
      for (int j = i; j < CMAX; j += i) rc[j] = i;
    }
    cnt[i] = 0;
    pk[i] = 1;
    int k = i;
    while (k > 1) {
      int q = rc[k];
      cnt[i]++;
      int ct = 0;
      while (k % q == 0) {
        k /= q;
        ct ^= 1;
      }
      if (ct) pk[i] *= q;
    }
  }
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    qr[r].push_back(make_pair(l, i));
  }
  for (int i = 0; i < 15; i++) best[i] = -1;
  for (int i = 0; i < CMAX; i++) {
    for (int j = 0; j < 8; j++) lst[i][j] = -1;
  }
  for (int i = 0; i < n; i++) {
    relax(a[i], i);
    sort(qr[i].begin(), qr[i].end());
    int an = 0;
    for (auto v : qr[i]) {
      while (an < 14 && best[an] < v.first) an++;
      ans[v.second] = an;
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}