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 maxn = 2e5 + 20;
const int maxq = 1e6 + 20 + maxn;
const int maxa = 6e6 + 20;
int l[maxq], r[maxq], a[maxn], ans[maxq];
int p[maxa], id[maxa];
bool has[maxa];
vector<int> query[maxn];
vector<pair<int, int> > shit[maxa];
int fen[maxn];
void update(int p, int val) {
  for (p = maxn - 10 - p; p < maxn; p += ((p) & (-(p))))
    fen[p] = min(fen[p], val);
}
int get(int p) {
  int res = 1e9;
  for (p = maxn - 10 - p; p > 0; p -= ((p) & (-(p)))) res = min(res, fen[p]);
  return res;
}
void add(int x, int ind) {
  vector<int> tmp;
  int y = id[x];
  while (y > 1) {
    tmp.push_back(p[y]);
    y /= p[y];
  }
  int m = tmp.size();
  for (int mask = 0; mask < (1 << m); mask++) {
    int mul = 1, t = 0;
    for (int i = 0; i < m; i++)
      if ((((mask) >> (i)) & 1)) mul *= tmp[i], t++;
    for (auto ker : shit[mul]) update(ker.second, m + ker.first - 2 * t);
    while (!shit[mul].empty() && shit[mul].back().first >= m)
      shit[mul].pop_back();
    shit[mul].push_back(make_pair(m, ind));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i], has[a[i]] = 1;
  fill(id, id + maxa, 1);
  memset(fen, 63, sizeof fen);
  for (int i = 2; i < maxa; i++)
    if (!p[i])
      for (int j = i; j < maxa; j += i) {
        p[j] = i;
        if (has[j]) {
          int cur = 0, now = j;
          while (now % i == 0) now /= i, cur ^= 1;
          if (cur) id[j] *= i;
        }
      }
  for (int i = 0; i < q; i++) {
    cin >> l[i] >> r[i];
    l[i]--;
    r[i]--;
    query[r[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    add(a[i], i);
    for (auto ind : query[i]) ans[ind] = get(l[ind]);
  }
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
}