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 = 200100;
const int N0 = 5100000;
bool isnotprime[N0];
int prime[N0], cnt, g[N0], f[N0], d[N0], ll[N0];
void get_prime() {
  ll[1] = 1, d[1] = 0, g[1] = 1;
  for (int i = 2; i <= 5032107; i++) {
    if (!isnotprime[i])
      prime[++cnt] = i, f[i] = 1, d[i] = 1, g[i] = i, ll[i] = i;
    for (int j = 1; j <= cnt && i * prime[j] <= 5032107; j++) {
      isnotprime[i * prime[j]] = true;
      if (i % prime[j]) {
        f[i * prime[j]] = 1, d[i * prime[j]] = d[i] + 1;
        g[i * prime[j]] = g[i] * prime[j], ll[i * prime[j]] = prime[j];
      } else {
        f[i * prime[j]] = f[i] + 1;
        if (f[i * prime[j]] & 1)
          g[i * prime[j]] = g[i] * prime[j];
        else
          g[i * prime[j]] = g[i] / prime[j];
        break;
      }
    }
  }
}
int dd[N], cd;
void dfs(int x, int s) {
  if (ll[x] == x) {
    dd[++cd] = s, dd[++cd] = x * s;
    return;
  }
  dfs(x / ll[x], s);
  dfs(x / ll[x], s * ll[x]);
}
vector<pair<int, int> > t0[N0], q0[N];
int a[N], c[N], ans[N0];
int main() {
  get_prime();
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = g[a[i]];
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    q0[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; i++) {
    cd = 0;
    dfs(a[i], 1);
    for (int j = 1; j <= cd; j++)
      if (t0[dd[j]].size())
        for (int k = 0; k <= t0[dd[j]].size() - 1; k++) {
          int tmp = t0[dd[j]][k].first + d[a[i] / dd[j]];
          c[tmp] = max(c[tmp], t0[dd[j]][k].second);
        }
    if (q0[i].size())
      for (int j = 0; j <= q0[i].size() - 1; j++)
        for (int k = 0; k <= 16; k++)
          if (c[k] >= q0[i][j].first) {
            ans[q0[i][j].second] = k;
            break;
          }
    for (int j = 1; j <= cd; j++) {
      while (t0[dd[j]].size() &&
             t0[dd[j]][t0[dd[j]].size() - 1].first >= d[a[i] / dd[j]])
        t0[dd[j]].pop_back();
      t0[dd[j]].push_back(make_pair(d[a[i] / dd[j]], i));
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}