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 N = 200005, MXA = 5090006, MXQ = 1090006, K = 9;
int n, q, A[N], R[MXQ], Rs[MXQ], LP[MXA], Frst[MXA][K], Mnr[K * 2];
vector<pair<int, int> > D[N];
vector<int> Q[N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 2; i < MXA; i++)
    if (!LP[i])
      for (int j = i; j < MXA; j += i) LP[j] = i;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    vector<int> Pr;
    while (a > 1) {
      if (Pr.size() && Pr.back() == LP[a])
        Pr.pop_back();
      else
        Pr.push_back(LP[a]);
      a /= LP[a];
    }
    A[i] = 1;
    for (int p : Pr) A[i] *= p;
    for (int mask = 0; mask < (1 << Pr.size()); mask++) {
      a = 1;
      for (int i = 0; i < Pr.size(); i++)
        if (mask >> i & 1) a *= Pr[i];
      D[i].push_back({a, (int)Pr.size() - __builtin_popcount(mask)});
    }
  }
  for (int i = 1; i <= q; i++) {
    int l;
    scanf("%d%d", &l, &R[i]);
    Q[l].push_back(i);
  }
  memset(Mnr, 63, sizeof(Mnr));
  memset(Frst, 63, sizeof(Frst));
  for (int i = n; i; i--) {
    for (auto X : D[i]) {
      int d = X.first;
      int c = X.second;
      for (int j = 0; j < K; j++) Mnr[c + j] = min(Mnr[c + j], Frst[d][j]);
      Frst[d][c] = i;
    }
    for (int j : Q[i])
      for (int h = 0; h < K * 2; h++)
        if (Mnr[h] <= R[j]) {
          Rs[j] = h;
          break;
        }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", Rs[i]);
  return 0;
}