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 = 5100000;
int n, q;
int a[200010];
struct query {
  int l, r, num;
} Q[1200000];
int h[maxn][8], rst[16];
vector<int> fac;
namespace EulerSieve {
const int N = 5050000;
int prime[maxn >> 3], flag[maxn], minn[maxn], num;
void getprime() {
  flag[1] = 1;
  minn[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!flag[i]) {
      minn[i] = i;
      prime[++num] = i;
    }
    for (int j = 1; j <= num && i * prime[j] <= N; j++) {
      flag[i * prime[j]] = 1;
      minn[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
int SimpleOut(int p) {
  int cnt = 0, dt = minn[p];
  int ans = 1;
  while (p != 1) {
    while (minn[p] == dt) {
      p /= dt;
      cnt ^= 1;
    }
    if (cnt) ans *= dt;
    dt = minn[p];
    cnt = 0;
  }
  return ans;
}
}  // namespace EulerSieve
void read() {
  scanf("%d%d", &n, &q);
  EulerSieve::getprime();
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = EulerSieve::SimpleOut(x);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &Q[i].l, &Q[i].r);
    Q[i].num = i;
  }
  sort(Q + 1, Q + q + 1, [](query u, query v) { return u.r < v.r; });
}
void dfs(int now, int chs, int tot, int dr, int hhh) {
  if (now == fac.size()) {
    if (dr == 0) {
      int rem = fac.size() - chs;
      int flag = 0;
      for (int i = 0; i <= 7; i++)
        if (h[tot][i] != 0) rst[rem + i] = max(rst[rem + i], h[tot][i]);
    } else {
      h[tot][fac.size() - chs] = hhh;
    }
  } else {
    dfs(now + 1, chs, tot, dr, hhh);
    dfs(now + 1, chs + 1, tot * fac[now], dr, hhh);
  }
}
int ans[maxn];
void work() {
  int j = 1;
  using EulerSieve::minn;
  for (int i = 1; i <= n; i++) {
    fac.clear();
    int p = a[i];
    if (p != 0)
      while (p != 1) {
        fac.push_back(minn[p]);
        p /= minn[p];
      }
    dfs(0, 0, 1, 0, i);
    dfs(0, 0, 1, 1, i);
    while (j <= q && Q[j].r == i) {
      for (int i = 0; i <= 14; i++)
        if (rst[i] >= Q[j].l) {
          ans[Q[j].num] = i;
          break;
        }
      j++;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
int main() {
  read();
  work();
  return 0;
}