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 nmax = 194598;
const int qmax = 1049658;
const int valmax = 5032107;
const int distmax = 8;
const int ansmax = 16;
struct Query {
  int stanga, indice;
};
int n, q;
int v[1 + nmax];
vector<Query> query[1 + nmax];
int best[1 + ansmax];
int dp[1 + valmax][1 + distmax];
vector<int> divs;
vector<int> nrFact;
int rasp[1 + qmax];
void citire() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 1; i <= q; i++) {
    int st, dr;
    cin >> st >> dr;
    query[dr].push_back({st, i});
  }
}
vector<int> descompunere(int x) {
  vector<int> fact;
  int d = 2, e, lim = sqrt(x);
  while (d <= lim && x > 1) {
    e = 0;
    while (x % d == 0) {
      e++;
      x /= d;
    }
    if (e % 2 == 1) {
      fact.push_back(d);
    }
    d++;
  }
  if (x > 1) {
    fact.push_back(x);
  }
  return fact;
}
int divizori(int x) {
  vector<int> fact = descompunere(x);
  divs.clear();
  nrFact.clear();
  divs.push_back(1);
  nrFact.push_back(1);
  for (int mask = 1; mask < (1 << fact.size()); mask++) {
    int cnt = 1, val = 1;
    for (int k = 0; k < fact.size(); ++k) {
      if (mask & (1 << k)) {
        cnt++;
        val *= fact[k];
      }
    }
    divs.push_back(val);
    nrFact.push_back(cnt);
  }
  return 1 + fact.size();
}
void solve() {
  for (int i = 1; i <= n; i++) {
    int nrTotalFactori = divizori(v[i]);
    for (int d = 0; d < divs.size(); d++) {
      int gcd = divs[d], dist = nrTotalFactori - nrFact[d];
      for (int k = 0; k <= distmax; k++) {
        best[k + dist] = max(best[k + dist], dp[gcd][k]);
      }
      dp[gcd][dist] = i;
    }
    for (int j = 0; j < query[i].size(); j++) {
      for (int rez = 0; rez <= ansmax; rez++) {
        if (query[i][j].stanga <= best[rez]) {
          rasp[query[i][j].indice] = rez;
          break;
        }
      }
    }
  }
}
void afis() {
  for (int i = 1; i <= q; i++) {
    cout << rasp[i] << '\n';
  }
}
int main() {
  citire();
  solve();
  afis();
  return 0;
}