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 = 6000000, N = 212345;
int br[maxn][10];
int p[maxn];
int n, q;
vector<int> fpr(int x) {
  vector<int> ret;
  while (p[x]) {
    ret.push_back(p[x]);
    x /= p[x];
  }
  if (x > 1) ret.push_back(x);
  sort(begin(ret), end(ret));
  vector<int> rret;
  for (int i = 0; i < ret.size();) {
    int j = i;
    while (j < ret.size() && ret[j] == ret[i]) j++;
    if ((j - i) & 1) rret.push_back(ret[i]);
    i = j;
  }
  return rret;
}
vector<int> vp[N];
int t[2 * N];
void update(int p, int v) {
  p += n;
  t[p] = min(t[p], v);
  while (p > 1) {
    t[p >> 1] = min(t[p], t[p ^ 1]);
    p >>= 1;
  }
}
int query(int l, int r) {
  int ans = 0x3f3f3f3f;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans = min(ans, t[l++]);
    if (r & 1) ans = min(ans, t[--r]);
  }
  return ans;
}
int ans[2123123];
vector<pair<int, int>> qr[N];
int main() {
  memset(t, 0x3f, sizeof t);
  for (int i = 2; i < maxn; i++) {
    if (!p[i]) {
      for (int j = i + i; j < maxn; j += i) {
        p[j] = i;
      }
    }
  }
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    vp[i] = fpr(x);
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--, r--;
    qr[r].emplace_back(l, i);
  }
  for (int i = 0; i < n; i++) {
    for (int m = 0; m < (1 << vp[i].size()); m++) {
      int mask = 1;
      for (int j = 0; j < vp[i].size(); j++) {
        if ((m >> j) & 1) mask *= vp[i][j];
      }
      for (int f = 0; f < 8; f++) {
        int cost = vp[i].size() - __builtin_popcount(m) + f;
        if (br[mask][f]) {
          update(br[mask][f] - 1, cost);
        }
      }
      int f = vp[i].size() - __builtin_popcount(m);
      br[mask][f] = max(br[mask][f], i + 1);
    }
    for (auto& qry : qr[i]) {
      int l, id;
      tie(l, id) = qry;
      ans[id] = query(l, i + 1);
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}