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 + 5, U = 51e5;
int mn[U];
int n, q, a[N], x, y, i, j, k, l, r, lb[199], cnt[199], buc[19], lst[U][9],
    ans[N * 6];
vector<pair<int, int>> qu[N];
inline void upi(int& a, const int& b) { a > b ? a = b : 0; }
int main() {
  for (i = 1; i < 199; ++i)
    lb[i] = i & 1 ? 0 : lb[i >> 1] + 1, cnt[i] = cnt[i >> 1] + (i & 1);
  for (i = 2; i < U; ++i)
    if (!mn[i])
      for (j = i; j < U; j += i)
        if (!mn[j]) mn[j] = i;
  memset(buc, 1, sizeof buc);
  memset(lst, 1, sizeof lst);
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  for (i = 1; i <= q; ++i) scanf("%d%d", &l, &r), qu[l].push_back({r, i});
  for (i = n; i; --i) {
    static int pr[19], pi[199];
    int xb = 0;
    for (x = a[i]; x > 1;) {
      int z = mn[x], c = 0;
      for (; x % z == 0; x /= z, ++c)
        ;
      if (c & 1) pr[xb++] = z;
    }
    *pi = 1;
    for (j = 1; j < 1 << xb; ++j) pi[j] = pi[j ^ (1 << lb[j])] * pr[lb[j]];
    for (j = 0; j < 1 << xb; ++j)
      for (k = cnt[j]; k <= 7; ++k)
        upi(buc[xb + k - cnt[j] * 2], lst[pi[j]][k]);
    for (j = 0; j < 1 << xb; ++j) lst[pi[j]][xb] = i;
    for (auto u : qu[i]) {
      tie(y, x) = u;
      for (k = 0; buc[k] > y; ++k)
        ;
      ans[x] = k;
    }
  }
  for (i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}