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, M = 2e6, P = 6e6;
inline int gi() {
  char ch = getchar();
  int q = 0, x = 0;
  while (ch < '0' || ch > '9') q = (ch == '-' ? 1 : q), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return q ? (-x) : x;
}
int a[N], p[N][8], t[N], ans[M], L[M], dp[P][10], f[20];
vector<int> Q[N];
int main() {
  int n = gi(), q = gi(), s, op;
  for (int i = 1; i <= n; ++i) {
    s = gi();
    for (int j = 2; j * j <= s; ++j) {
      op = 0;
      while (s % j == 0) op ^= 1, s /= j;
      if (op) p[i][t[i]++] = j;
    }
    if (s > 1) p[i][t[i]++] = s;
  }
  for (int i = 1; i <= q; ++i)
    scanf("%d%d", &s, &op), Q[op].push_back(i), L[i] = s;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (1 << t[i]); ++j) {
      int cnt = 0;
      s = 1;
      for (int k = 0; k < t[i]; ++k)
        if (j >> k & 1)
          s *= p[i][k];
        else
          ++cnt;
      for (int k = 0; k < 8; ++k) f[k + cnt] = max(f[k + cnt], dp[s][k]);
    }
    for (int j = 0; j < Q[i].size(); ++j)
      for (int k = 0; k < 16; ++k)
        if (f[k] >= L[Q[i][j]]) {
          ans[Q[i][j]] = k;
          break;
        }
    for (int j = 0; j < (1 << t[i]); ++j) {
      int cnt = 0;
      s = 1;
      for (int k = 0; k < t[i]; ++k)
        if (j >> k & 1)
          s *= p[i][k];
        else
          ++cnt;
      dp[s][cnt] = i;
    }
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
  return 0;
}