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;
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  int m = *max_element(a.begin(), a.end());
  vector<int> primes, minp(m + 1);
  for (int i = 2; i <= m; ++i) {
    if (!minp[i]) {
      minp[i] = i;
      primes.push_back(i);
    }
    for (auto p : primes) {
      if (i * p > m) {
        break;
      }
      minp[i * p] = p;
      if (i % p == 0) {
        break;
      }
    }
  }
  auto factorization = [&](int x) {
    vector<pair<int, int>> result;
    vector<int> primes;
    while (x != 1) {
      int counter = 0, p = minp[x];
      while (x % p == 0) {
        x /= p;
        ++counter;
      }
      if (counter & 1) {
        primes.push_back(p);
      }
    }
    function<void(int, int, int)> dfs = [&](int x, int product, int dist) {
      if (x == primes.size()) {
        result.emplace_back(product, dist);
      } else {
        dfs(x + 1, product * primes[x], dist);
        dfs(x + 1, product, dist + 1);
      }
    };
    dfs(0, 1, 0);
    return result;
  };
  vector<vector<int>> answer(n, vector<int>(15));
  vector<vector<int>> right(m + 1, vector<int>(8, n));
  for (int i = n - 1; ~i; --i) {
    for (int j = 0; j < 15; ++j) {
      answer[i][j] = n;
    }
    vector<pair<int, int>> all = factorization(a[i]);
    for (auto p : all) {
      for (int j = 0; j < 8; ++j) {
        answer[i][j + p.second] =
            min(answer[i][j + p.second], right[p.first][j]);
      }
    }
    for (auto p : all) {
      right[p.first][p.second] = i;
    }
    for (int j = 1; j < 15; ++j) {
      answer[i][j] = min(answer[i][j], answer[i][j - 1]);
    }
  }
  vector<int> l(q), r(q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d", &l[i], &r[i]);
    --l[i];
    --r[i];
  }
  vector<int> result(q, 15);
  for (int i = 14; ~i; --i) {
    vector<int> to(n);
    for (int j = 0; j < n; ++j) {
      to[j] = answer[j][i];
    }
    for (int j = n - 1; j; --j) {
      to[j - 1] = min(to[j - 1], to[j]);
    }
    for (int j = 0; j < q; ++j) {
      if (to[l[j]] <= r[j]) {
        result[j] = i;
      }
    }
  }
  for (int i = 0; i < q; ++i) {
    printf("%d\n", result[i]);
  }
  return 0;
}