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, max_value = 5032108;
void cmin(int& x, int y) {
  if (x > y) {
    x = y;
  }
}
int n, q, a[N], minp[max_value], ans[N][16], min_step[max_value][8];
bool is_prime[max_value];
vector<int> primes;
void sieve() {
  memset(is_prime, true, sizeof is_prime);
  for (int i = 2; i <= max_value; ++i) {
    if (!minp[i]) {
      minp[i] = i;
      primes.push_back(i);
    }
    for (auto v : primes) {
      int d = i * v;
      if (d > max_value) {
        break;
      }
      minp[d] = v;
      if (i % v == 0) {
        break;
      }
    }
  }
}
int main() {
  sieve();
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n + 1; ++i) {
    for (int j = 0; j <= 14; ++j) {
      ans[i][j] = n + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  auto get_sets = [&](int value) {
    vector<int> prime_e;
    vector<pair<int, int>> result;
    while (value ^ 1) {
      int pri = minp[value], total = 0;
      for (; value % pri == 0; ++total, value /= pri)
        ;
      if (total & 1) {
        prime_e.push_back(pri);
      }
    }
    function<void(int, int, int)> dfs = [&](int x, int v, int dist) {
      if (x == prime_e.size()) {
        result.emplace_back(v, dist);
      } else {
        dfs(x + 1, v * prime_e[x], dist);
        dfs(x + 1, v, dist + 1);
      }
    };
    dfs(0, 1, 0);
    return result;
  };
  memset(min_step, -1, sizeof min_step);
  for (int i = n; i; --i) {
    vector<pair<int, int>> all = get_sets(a[i]);
    for (auto p : all) {
      for (int j = 0; j <= 7; ++j) {
        if (~min_step[p.first][j]) {
          cmin(ans[i][p.second + j], min_step[p.first][j]);
        }
      }
    }
    for (auto p : all) {
      min_step[p.first][p.second] = i;
    }
    for (int j = 0; j <= 14; ++j) {
      cmin(ans[i][j], ans[i + 1][j]);
    }
  }
  for (int i = 1; i <= q; ++i) {
    int l, r, result = 0;
    scanf("%d%d", &l, &r);
    for (;; ++result) {
      if (ans[l][result] <= r) {
        break;
      }
    }
    printf("%d\n", result);
  }
  return 0;
}