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 = 200000, M = 5100000;
int n, cq;
vector<int> pr[N + 9], ql[N + 9], qid[N + 9];
void into() {
  scanf("%d%d", &n, &cq);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    int t = x;
    for (int j = 2; j * j <= x; ++j) {
      if (t % j) continue;
      int c = 0;
      for (; t % j == 0; t /= j) c ^= 1;
      if (c) pr[i].push_back(j);
    }
    if (t) pr[i].push_back(t);
  }
  for (int i = 1; i <= cq; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    ql[r].push_back(l);
    qid[r].push_back(i);
  }
}
int p[M + 9][9], pos[20];
int ans[M + 9];
void Get_ans() {
  for (int i = 1; i <= n; ++i) {
    int c = pr[i].size();
    for (int s = 0; s < 1 << c; ++s) {
      int x = 1, y = 0;
      for (int j = 0; j < c; ++j) s >> j& 1 ? x *= pr[i][j] : ++y;
      for (int j = 0; j <= 7; ++j) pos[j + y] = max(pos[j + y], p[x][j]);
    }
    for (int vs = ql[i].size(), j = 0; j < vs; ++j) {
      int t = 0;
      for (; t <= 15 && pos[t] < ql[i][j]; ++t)
        ;
      ans[qid[i][j]] = t;
    }
    for (int s = 0; s < 1 << c; ++s) {
      int x = 1, y = 0;
      for (int j = 0; j < c; ++j) s >> j& 1 ? x *= pr[i][j] : ++y;
      p[x][y] = i;
    }
  }
}
void work() { Get_ans(); }
void outo() {
  for (int i = 1; i <= cq; ++i) printf("%d\n", ans[i]);
}
int main() {
  int T = 1;
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}