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 A = 5032108;
const int N = 194598;
const int Q = 1049658;
int last[8][A];
int a[N];
int l[Q], r[Q];
int n, q;
class Primes {
 public:
  Primes(int limit) {
    isPrime = new bool[limit];
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i < limit; i++) isPrime[i] = true;
    for (int i = 2; i < limit; i++)
      if (isPrime[i]) {
        primes.push_back(i);
        if (((long long)i) * i < limit)
          for (int j = i * i; j < limit; j += i) isPrime[j] = false;
      }
  }
  ~Primes() { delete[] isPrime; }
  vector<int> primes;
  bool* isPrime;
};
bool cmp(int i, int j) { return r[i] < r[j]; }
Primes p(2244);
int fact[7];
int factorize(int a) {
  int count = 0;
  for (int i = 0; i < p.primes.size() && p.primes[i] * p.primes[i] <= a; i++)
    if (a % p.primes[i] == 0) {
      bool odd = false;
      while (a % p.primes[i] == 0) {
        a /= p.primes[i];
        odd = !odd;
      }
      if (odd) fact[count++] = p.primes[i];
    }
  if (a > 1) fact[count++] = a;
  return count;
}
int main() {
  cin.sync_with_stdio(0);
  cout.sync_with_stdio(0);
  cin.tie(0);
  for (int i = 0; i < 8; i++)
    for (int j = 0; j < A; j++) last[i][j] = -1;
  cin >> n >> q;
  vector<int> ind(q);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) {
    cin >> l[i] >> r[i];
    l[i]--;
    r[i]--;
    ind[i] = i;
  }
  sort(ind.begin(), ind.end(), cmp);
  vector<int> ans(q);
  int best[15];
  for (int i = 0; i < 15; i++) best[i] = -1;
  int curq = 0;
  for (int i = 0; i < n; i++) {
    int count = factorize(a[i]);
    int aa = 1;
    for (int j = 0; j < count; j++) aa *= fact[j];
    int limit = 1 << count;
    for (int j = 0; j < limit; j++) {
      int delcount = 0;
      int rem = aa;
      for (int bit = 1, k = 0; bit < limit; bit <<= 1, k++)
        if (j & bit) {
          delcount++;
          rem /= fact[k];
        }
      for (int k = 0; k < 8; k++)
        if (last[k][rem] != -1)
          best[delcount + k] = max(best[delcount + k], last[k][rem]);
    }
    for (int j = 0; j < limit; j++) {
      int delcount = 0;
      int rem = aa;
      for (int bit = 1, k = 0; bit < limit; bit <<= 1, k++)
        if (j & bit) {
          delcount++;
          rem /= fact[k];
        }
      last[delcount][rem] = i;
    }
    while (curq < q && r[ind[curq]] == i) {
      int j = ind[curq++];
      ans[j] = 14;
      for (int k = 0; k < 15; k++)
        if (best[k] >= l[j]) {
          ans[j] = k;
          break;
        }
    }
  }
  for (int x : ans) cout << x << "\n";
  return 0;
}