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, V = 51e5;
int n, q, f[N][16], bc[1 << 8 | 10];
int bt[V][8];
int pri[V / 10], pnum, npr[V];
inline void U(int &x, int y) { x < y ? x = y : 0; }
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 2; i < V; ++i)
    if (!npr[i]) {
      pri[++pnum] = i;
      for (int j = i + i; j < V; j += i) npr[j] = 1;
    }
  for (int i = 0; i < 1 << 8; ++i) bc[i] = bc[i >> 1] + (i & 1);
  for (int id = 1, x; id <= n; ++id) {
    scanf("%d", &x);
    int top = 0, sta[8];
    for (int i = 1; pri[i] * pri[i] <= x; ++i) {
      int nd = 0;
      while (x % pri[i] == 0) x /= pri[i], nd ^= 1;
      if (nd) sta[top++] = pri[i];
    }
    if (x > 1) sta[top++] = x;
    for (int i = 0; i < 1 << top; ++i) {
      int ans = 1;
      for (int j = 0; j < top; ++j)
        if (i >> j & 1) ans *= sta[j];
      for (int j = 0; j < 8; ++j) U(f[id][j + top - bc[i]], bt[ans][j]);
      bt[ans][top - bc[i]] = id;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j < 16; ++j) U(f[i][j], f[i][j - 1]);
    for (int j = 0; j < 16; ++j) U(f[i][j], f[i - 1][j]);
  }
  for (int id = 1, l, r; id <= q; ++id) {
    scanf("%d%d", &l, &r);
    for (int i = 0; i < 16; ++i)
      if (f[r][i] >= l) {
        printf("%d\n", i);
        break;
      }
  }
  return 0;
}