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 = (int)2e5 + 50, K = 8, Q = 1050000;
int n, q;
int a[N];
vector<pair<int, int> > qs[N];
const int M = 5032107 + 50;
int p[M];
int mu[M], pm[M], pms;
void get_mobius_and_sieve() {
  mu[1] = 1;
  for (int i = 0; i < M; i++) p[i] = i;
  pms = 0;
  for (int i = 2; i < M; i++) {
    if (p[i] == i) {
      pm[pms++] = i;
      mu[i] = -1;
    }
    for (int j = 0; j < pms && i * pm[j] < M; j++) {
      p[i * pm[j]] = pm[j];
      if (i % pm[j] == 0) {
        mu[i * pm[j]] = 0;
        break;
      }
      mu[i * pm[j]] = -mu[i];
    }
  }
}
vector<pair<int, int> > get_divs(int x) {
  vector<int> d;
  vector<pair<int, int> > res;
  while (x > 1) d.push_back(p[x]), x /= p[x];
  int b = d.size();
  for (int i = 0; i < (1 << b); i++) {
    int cur = 1, cnt = b;
    for (int j = 0; j < b; j++)
      if (i & (1 << j)) cur *= d[j], cnt--;
    res.push_back({cur, cnt});
  }
  return res;
}
int to[M];
int rt[M][K];
int best[K * 2];
int res[Q];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  get_mobius_and_sieve();
  for (int i = 0; i < M; i++) to[i] = i;
  for (int i = 1; i * i < M; i++) {
    for (int j = 1; j * i * i < M; j++) {
      to[j * i * i] = min(to[j * i * i], j);
    }
  }
  memset(rt, -1, sizeof(rt));
  memset(best, -1, sizeof(best));
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i], a[i] = to[a[i]];
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    qs[r].push_back({l, i});
  }
  for (int i = 0; i < n; i++) {
    auto dv = get_divs(a[i]);
    for (auto p : dv) {
      for (int j = 0; j < K; j++)
        best[p.second + j] = max(best[p.second + j], rt[p.first][j]);
    }
    for (auto p : dv) rt[p.first][p.second] = i;
    for (auto qry : qs[i]) {
      for (int j = 0; j < 2 * K; j++) {
        if (best[j] >= qry.first) {
          res[qry.second] = j;
          break;
        }
      }
    }
  }
  for (int i = 0; i < q; i++) cout << res[i] << "\n";
}