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 = 6e6 + 5, Q = 2e6 + 5;
int n, q, f[N][8], g[20], a[N], t[1 << 8], ans[N];
vector<pair<int, int> > qry[Q];
vector<int> trans(int x) {
  vector<int> d, res;
  for (int i = 2; i * i <= x; ++i) {
    if (x % i) continue;
    int c = 0;
    while (x % i == 0) x /= i, c ^= 1;
    if (c) d.push_back(i);
  }
  if (x > 1) d.push_back(x);
  return d;
}
void getmax(int &x, int y) { x = max(x, y); }
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= int(n); ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= int(q); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    qry[r].push_back({l, i});
  }
  for (int i = 1; i <= int(n); ++i) {
    auto v = trans(a[i]);
    int lim = 1 << v.size();
    t[0] = 1;
    for (int s = 0; s <= int(lim - 1); ++s) {
      if (s) t[s] = t[s ^ (s & -s)] * v[__builtin_ctz(s)];
      int x = t[s];
      int l1 = v.size() - __builtin_popcount(s);
      for (int j = 0; j <= int(7); ++j) {
        getmax(g[j + l1], f[x][j]);
      }
    }
    for (int s = 0; s <= int(lim - 1); ++s) {
      int x = t[s];
      int l1 = v.size() - __builtin_popcount(s);
      getmax(f[x][l1], i);
    }
    for (auto qr : qry[i]) {
      for (int j = 0; j <= int(19); ++j)
        if (qr.first <= g[j]) {
          ans[qr.second] = j;
          break;
        }
    }
  }
  for (int i = 1; i <= int(q); ++i) printf("%d\n", ans[i]);
  return 0;
}