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;
constexpr int MAX_N = 200200;
constexpr int MAX = 5100000;
int n, m;
int arr[MAX_N];
int sieve[MAX];
int last[MAX][8];
int ans[MAX_N][15];
void build_sieve() {
  for (int i = 2; i < MAX; ++i) {
    if (sieve[i] == 0) {
      for (int j = i; j < MAX; j += i) {
        sieve[j] = i;
      }
    }
  }
}
vector<int> factors(int x) {
  map<int, int> s;
  while (x != 1) {
    s[sieve[x]] = !s[sieve[x]];
    x /= sieve[x];
  }
  vector<int> ans;
  for (pair<int, int> p : s) {
    if (p.second == 1) {
      ans.push_back(p.first);
    }
  }
  sort(ans.begin(), ans.end());
  return ans;
}
int main() {
  build_sieve();
  scanf(" %d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf(" %d", &arr[i]);
  }
  for (int x = 0; x < MAX; ++x) {
    for (int y = 0; y <= 7; ++y) {
      last[x][y] = n + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int w = 0; w <= 14; ++w) {
      ans[i][w] = n + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    vector<int> f = factors(arr[i]);
    for (int mask = 0; mask < (1 << (int)f.size()); ++mask) {
      int prod = 1;
      for (int j = 0; j < (int)f.size(); ++j) {
        if ((1 << j) & mask) prod *= f[j];
      }
      int work = (int)f.size() - (int)__builtin_popcount(mask);
      for (int fut = 0; fut <= 7; ++fut) {
        ans[i][work + fut] = min(ans[i][work + fut], last[prod][fut]);
      }
      last[prod][work] = i;
    }
  }
  for (int i = n - 1; i >= 1; --i) {
    for (int w = 0; w <= 14; ++w) {
      ans[i][w] = min(ans[i][w], ans[i + 1][w]);
    }
  }
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf(" %d %d", &l, &r);
    int cur = 15;
    for (int w = 14; w >= 0; --w) {
      if (ans[l][w] <= r) cur = w;
    }
    assert(cur < 15);
    printf("%d\n", cur);
  }
  return 0;
}