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 = 2e5, V = 51e5, M = 15, D = 8, Q = 11e5, S = 3000;
vector<pair<int, int>> qr[N];
bool comp[S];
int pr[S], ans[Q], dp[V][D], best[M], a[N], ct[V], sz[128];
int cp = 0;
int main() {
  for (int i = 2; i < S; i++) {
    if (!comp[i]) {
      pr[cp++] = i;
      for (int j = i * i; j < S; j += i) {
        comp[j] = 1;
      }
    }
  }
  sz[0] = 0;
  for (int i = 1; i < 128; i++) sz[i] = sz[i ^ (i & -i)] + 1;
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < V; i++) {
    for (int j = 0; j < D; j++) {
      dp[i][j] = -1;
    }
  }
  for (int i = 0; i < M; i++) best[i] = -1;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    qr[r - 1].push_back({l - 1, i});
  }
  for (int i = 0; i < n; i++) {
    vector<int> d;
    int t = a[i];
    for (int j = 0; pr[j] * pr[j] <= t; j++) {
      int c = 0;
      while (t % pr[j] == 0) {
        c ^= 1;
        t /= pr[j];
      }
      if (c) d.push_back(pr[j]);
    }
    if (t > 1) d.push_back(t);
    vector<int> g(1 << d.size());
    g[0] = 1;
    for (int j = 0; j < d.size(); j++) {
      g[1 << j] = d[j];
    }
    for (int j = 1; j < (1 << d.size()); j++) {
      g[j] = g[j & -j] * g[j ^ (j & -j)];
      ct[g[j]] = sz[j];
    }
    ct[a[i]] = d.size();
    for (int j : g) {
      for (int k = 0; k < M; k++) {
        int c = k + ct[a[i]] - ct[j];
        best[c] = max(best[c], dp[j][k]);
      }
      dp[j][ct[a[i]] - ct[j]] = i;
    }
    for (auto p : qr[i]) {
      for (int k = 0; k < M; k++) {
        if (best[k] >= p.first) {
          ans[p.second] = k;
          break;
        }
      }
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}