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 = 5000000;
int prime[Maxn + 5], tot;
bool check[Maxn + 5];
int bit[1 << 8];
void pre() {
  for (int i = 2; i <= Maxn; i++) {
    if (!check[i]) prime[++tot] = i;
    for (int j = 1; j <= tot && i * prime[j] <= Maxn; j++) {
      check[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 0; i < (1 << 8); i++) bit[i] = bit[i >> 1] + (i & 1);
}
vector<int> vt[200005];
void getfact(int n, int id) {
  int t = n;
  for (int i = 1; i <= tot && prime[i] * prime[i] <= t; i++)
    if (n % prime[i] == 0) {
      bool v = 0;
      while (n % prime[i] == 0) {
        n /= prime[i];
        v ^= 1;
      }
      if (v) vt[id].push_back(prime[i]);
    }
  if (n > 1) vt[id].push_back(n);
}
int maxn[6000000][10], f[20];
void update(int id) {
  int len = vt[id].size();
  for (int i = 0; i < (1 << len); i++) {
    int s = 1;
    for (int j = 0; j < len; j++)
      if (i & (1 << j)) s *= vt[id][j];
    for (int j = 0; j < 10; j++)
      if (maxn[s][j])
        f[len + j - 2 * bit[i]] = max(f[len + j - 2 * bit[i]], maxn[s][j]);
    maxn[s][len] = id;
  }
}
struct Query {
  int l, r, id;
  Query() {}
  Query(int a, int b, int c) : l(a), r(b), id(c) {}
  bool operator<(const Query& b) const { return r < b.r; }
};
Query q[1100000];
int ans[1100000];
int main() {
  pre();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    getfact(x, i);
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    q[i] = Query(x, y, i);
  }
  sort(q + 1, q + m + 1);
  int r = 1;
  for (int i = 1; i <= n; i++) {
    update(i);
    while (r <= m && q[r].r <= i) {
      for (int j = 0; j < 20; j++)
        if (f[j] >= q[r].l) {
          ans[q[r].id] = j;
          break;
        }
      r++;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}