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>
const int big = 8;
const int lim = 5032107;
const int maxn = 1111111;
int n, q, D[lim + 17], W[lim + 17], ans[maxn], a[maxn], l[maxn], r[maxn],
    f[100], R[lim + 17][10];
std::vector<int> v[200005], rec[200005];
std::vector<int> work(int x) {
  std::vector<int> v;
  while (x != 1) {
    int d = D[x];
    v.push_back(d);
    while (x % d == 0) x /= d;
  }
  return v;
}
std::vector<int> process(int x) {
  std::vector<int> v = work(x), ans;
  int len = v.size();
  for (int S = 0; S < (1 << len); ++S) {
    int w = 1;
    for (int i = 0; i < len; ++i)
      if (S & (1 << i)) w *= v[i];
    ans.push_back(w);
  }
  return ans;
}
int magic(int x) {
  int y = 1;
  std::vector<int> v;
  while (x != 1) {
    int d = D[x];
    int c = 0;
    while (x % d == 0) {
      x /= d;
      c ^= 1;
    }
    if (c) y *= d;
  }
  return y;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 2; i <= lim; ++i) {
    if (D[i]) continue;
    for (int j = i; j <= lim; j += i) {
      D[j] = i;
      W[j]++;
    }
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    a[i] = magic(a[i]);
    v[i] = process(a[i]);
  }
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &l[i], &r[i]);
    rec[r[i]].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    for (int x : v[i]) {
      for (int j = 0; j <= big; ++j) {
        int sum = W[a[i]] + j - 2 * W[x];
        if (R[x][j]) {
          f[sum] = std::max(f[sum], R[x][j]);
        }
      }
      R[x][W[a[i]]] = i;
    }
    for (int x : rec[i]) {
      for (int j = 0; j <= 3 * big; ++j)
        if (f[j] >= l[x]) {
          ans[x] = j;
          break;
        }
    }
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
  return 0;
}