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 maxn = 200000, maxv = 6000000;
int n, q;
int prm[maxv + 10], pcnt, g[maxv + 10], c[maxv + 10], a[maxn + 10], acnt;
int f[maxv + 10][8], h[maxn + 10][15];
bool nprm[maxv + 10];
void pre() {
  for (int i = 2; i <= maxv; ++i) {
    if (!nprm[i]) {
      prm[++pcnt] = i;
      g[i] = i;
    }
    for (int j = 1; j <= pcnt && i * prm[j] <= maxv; ++j) {
      nprm[i * prm[j]] = 1;
      g[i * prm[j]] = prm[j];
      if (i % prm[j] == 0) break;
    }
  }
}
int main() {
  pre();
  scanf("%d%d", &n, &q);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    acnt = 0;
    for (int t = x; t > 1;) {
      int v = g[t], b = 0;
      while (t % v == 0) {
        t /= v;
        b ^= 1;
      }
      if (b) a[acnt++] = v;
    }
    for (int j = 0; j <= 14; ++j) h[i][j] = h[i - 1][j];
    for (int j = 0; j < 1 << acnt; ++j) {
      int v = 1, d = acnt;
      for (int k = 0; k < acnt; ++k)
        if (j >> k & 1) {
          v *= a[k];
          --d;
        }
      for (int k = 0; k <= 7; ++k) h[i][d + k] = max(h[i][d + k], f[v][k]);
    }
    for (int j = 0; j < 1 << acnt; ++j) {
      int v = 1, d = acnt;
      for (int k = 0; k < acnt; ++k)
        if (j >> k & 1) {
          v *= a[k];
          --d;
        }
      f[v][d] = i;
    }
  }
  while (q--) {
    int l, r, ans = 1e9;
    scanf("%d%d", &l, &r);
    for (int i = 0; i <= 14; ++i)
      if (l <= h[r][i]) ans = min(ans, i);
    printf("%d\n", ans);
  }
}