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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
int n, lpd[5100000];
int last[5100000][8];
int st[15][2 * 200010];
void update(int p, int nr, int pos) {
  if (st[nr][p += n] >= pos) return;
  for (st[nr][p] = pos; p > 1; p >>= 1)
    st[nr][p >> 1] = max(st[nr][p], st[nr][p ^ 1]);
}
int query(int nr, int l, int r) {
  int res;
  for (res = -1, l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = max(res, st[nr][l++]);
    if (r & 1) res = max(res, st[nr][--r]);
  }
  return res;
}
void process(int a, int i) {
  int p, e;
  vector<int> d;
  while (a > 1) {
    for (e = 0, p = lpd[a]; a % p == 0; ++e) a /= p;
    if (e & 1) d.push_back(p);
  }
  int n = d.size();
  for (int cfg = 0; cfg < (1 << n); ++cfg) {
    int j, val, s;
    for (s = 0, val = 1, j = 0; j < n; ++j)
      if (cfg & (1 << j)) val *= d[j], ++s;
    for (int nr = 0; nr <= 7; ++nr) update(i, n - s + nr, last[val][nr]);
    last[val][n - s] = i;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, q, a, l, r, ans;
  for (i = 1; i < 5100000; ++i) lpd[i] = i;
  for (i = 2; i < 5100000; ++i) {
    if (lpd[i] != i) continue;
    for (ll j = 1LL * i * i; j < 5100000; j += i) lpd[j] = min(lpd[j], i);
  }
  cin >> n >> q;
  memset(st, -1, sizeof st);
  memset(last, -1, sizeof last);
  for (i = 0; i < n; ++i) {
    cin >> a;
    process(a, i);
  }
  while (q--) {
    cin >> l >> r;
    for (ans = 0; ans <= 14; ++ans)
      if (query(ans, l - 1, r) >= l - 1) {
        cout << ans << '\n';
        break;
      }
  }
  return 0;
}