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 INF = 2147483647;
inline int read() {
  int neg = 1, num = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') neg = -1;
  for (; isdigit(c); c = getchar()) num = (num << 1) + (num << 3) + c - '0';
  return neg * num;
}
int n, Q, C, tot, a[200010], ans[15], lst[1100000], pri[200010 * 10],
    cnt[5100000], num[5100000], pos[5100000][8];
bool vis[5100000];
vector<int> d[200010];
vector<pair<int, int> > q[200010];
inline void sieve() {
  num[1] = vis[1] = 1;
  int maxn = 5032107;
  for (int i = 2; i <= maxn; i++) {
    if (!vis[i]) pri[++tot] = i, num[i] = i, cnt[i] = 1;
    for (int j = 1; j <= tot && pri[j] * i <= maxn; j++) {
      vis[pri[j] * i] = 1;
      num[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
      cnt[i * pri[j]] = cnt[i] + 1;
    }
  }
}
inline void divide(int x, int sum, int id) {
  if (x == 1) {
    d[id].push_back(sum);
    return;
  }
  divide(x / num[x], sum, id);
  divide(x / num[x], sum * num[x], id);
}
int main() {
  n = read();
  Q = read();
  sieve();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) {
    int res = 1;
    while (a[i] != 1) {
      int sum = 0, x = num[a[i]];
      while (a[i] % x == 0) sum++, a[i] /= x;
      if (sum & 1) res *= x;
    }
    a[i] = res;
    divide(a[i], 1, i);
    C = max(C, a[i]);
  }
  for (int i = 1; i <= Q; i++) {
    int l = read(), r = read();
    q[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; i++) {
    for (auto v : d[i]) {
      int sum = cnt[a[i]] - 2 * cnt[v];
      for (int j = 0; j <= 7; j++) ans[j + sum] = max(ans[j + sum], pos[v][j]);
      pos[v][cnt[a[i]]] = i;
    }
    for (auto v : q[i])
      for (int j = 14; j >= 0; j--)
        if (ans[j] >= v.first) lst[v.second] = j;
  }
  for (int i = 1; i <= Q; i++) printf("%d\n", lst[i]);
  return 0;
}