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;
int n;
int a[200000];
vector<int> as[200000];
int mx[5100000][8];
int st[200000][15];
void trans(int x) {
  int v = a[x];
  for (int i = 2; i * i <= v; ++i) {
    while (v % (i * i) == 0) v /= i * i;
    if (v % i == 0) as[x].push_back(i), v /= i;
  }
  if (v > 1) as[x].push_back(v);
}
void updateMx(int t, int i, int x, int c) {
  if (i == as[t].size()) {
    mx[x][c] = max(mx[x][c], t);
    return;
  }
  updateMx(t, i + 1, x, c);
  updateMx(t, i + 1, x * as[t][i], c - 1);
}
void calcSt(int t, int i, int x, int c) {
  if (i == as[t].size()) {
    for (i = 0; i <= 7; ++i) st[t][i + c] = max(st[t][i + c], mx[x][i]);
    return;
  }
  calcSt(t, i + 1, x, c);
  calcSt(t, i + 1, x * as[t][i], c - 1);
}
struct _qs {
  int l, r, i;
  void scan(int idx) {
    i = idx;
    cin >> l >> r;
  }
  bool operator<(const _qs &p) const { return r < p.r; }
} qs[1100000];
int ans[1100000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    trans(i);
    calcSt(i, 0, 1, as[i].size());
    for (int j = 1; j < 15; ++j) st[i][j] = max(st[i][j], st[i][j - 1]);
    updateMx(i, 0, 1, as[i].size());
  }
  for (int i = 0; i < q; ++i) qs[i].scan(i);
  sort(qs, qs + q);
  for (int i = 0, j = 1; i < q; ++i) {
    while (j <= qs[i].r) {
      for (int k = 0; k < 15; ++k) st[j][k] = max(st[j][k], st[j - 1][k]);
      ++j;
    }
    int s = 0, e = 15;
    while (s < e) {
      int m = (s + e) / 2;
      if (qs[i].l <= st[qs[i].r][m])
        e = m;
      else
        s = m + 1;
    }
    ans[qs[i].i] = s;
  }
  for (int i = 0; i < q; ++i) printf("%d\n", ans[i]);
  return 0;
}