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>
#pragma warning(disable : 4996)
const int bign = 200033;
const int bigm = 5032107;
int dp[bigm + 33][8];
int G[bign][17];
int minprime[bigm + 33], n, q;
int plist[9];
int main() {
  for (int i = 2; i <= bigm; i++) {
    if (!minprime[i]) {
      minprime[i] = i;
    } else {
      continue;
    }
    for (int j = i + i; j <= bigm; j += i) {
      if (!minprime[j]) {
        minprime[j] = i;
      }
    }
  }
  scanf("%d%d", &n, &q);
  for (int f = 1; f <= n; f++) {
    int x;
    scanf("%d", &x);
    int mtop = 0;
    while (x > 1) {
      int tmp = minprime[x];
      int tcnt = 0;
      while (x % tmp == 0) {
        x /= tmp;
        tcnt++;
      }
      if (tcnt & 1) {
        plist[mtop++] = tmp;
      }
    }
    for (int j = 0; j <= 14; j++) {
      G[f][j] = G[f - 1][j];
    }
    for (int i = 0; i < (1 << mtop); i++) {
      int tres = 1;
      int tcnt = 0;
      for (int j = 1, k = 0; j <= i; j <<= 1, k++) {
        if (i & j) {
          tres *= plist[k];
          tcnt++;
        }
      }
      for (int j = 0; j <= 7; j++) {
        if (dp[tres][j]) {
          G[f][j + mtop - tcnt] = std::max(G[f][j + mtop - tcnt], dp[tres][j]);
        }
      }
      dp[tres][mtop - tcnt] = f;
    }
  }
  for (int i = 1; i <= q; i++) {
    int lii, ri;
    scanf("%d%d", &lii, &ri);
    for (int j = 0; j <= 12; j++) {
      if (G[ri][j] >= lii) {
        printf("%d\n", j);
        break;
      }
    }
  }
}