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 = 194600;
const int LIM = 5032110;
const int Q = 1049660;
int n, q, a[N], large[LIM], f[N][16], dp[LIM][8], ans[Q];
vector<pair<int, int>> query[N];
vector<int> decomp(int x) {
  vector<int> v;
  while (x != 1) {
    int tmp = large[x], cnt = 0;
    while (x % tmp == 0) cnt++, x /= tmp;
    if (cnt & 1) v.push_back(tmp);
  }
  return v;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 2; i < LIM; i++) large[i] = i;
  for (int i = 2; i * i <= LIM; i++)
    if (large[i] == i) {
      for (int j = i * i; j < LIM; j += i)
        if (large[j] == j) large[j] = i;
    }
  memset(f, -1, sizeof(f));
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) {
    if (i > 0)
      for (int j = 0; j < 15; j++) f[i][j] = f[i - 1][j];
    vector<int> v = decomp(a[i]);
    int sz = v.size();
    for (int msk = 0; msk < (1 << sz); msk++) {
      int val = 1;
      for (int j = 0; j < sz; j++)
        if (msk & (1 << j)) val *= v[j];
      int tmp = v.size() - __builtin_popcount(msk);
      for (int k = 0; k < 8; k++)
        f[i][tmp + k] = f[i][tmp + k] > dp[val][k] ? f[i][tmp + k] : dp[val][k];
      dp[val][tmp] = i;
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    for (int i = 0; i < 15; i++)
      if (f[r][i] >= l) {
        cout << i << "\n";
        break;
      }
  }
  return 0;
}