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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int N = 5032110;
const int Q = 1049660;
int ans[15][Q], last[8][N];
int lp[N], arr[200009];
vector<int> mem;
int solve(int x) {
  mem.clear();
  int last = -1, cnt = 0;
  int res = 1;
  while (x > 1) {
    if (last != lp[x]) {
      if (~last) {
        if (cnt & 1) {
          mem.push_back(last);
          res *= last;
        }
      }
      cnt = 1;
    } else
      cnt++;
    last = lp[x];
    x /= lp[x];
  }
  if (cnt & 1) {
    res *= last;
    mem.push_back(last);
  }
  return res;
}
int main() {
  for (int i = 2; i < N; i++) {
    if (lp[i]) continue;
    for (int j = i; j < N; j += i)
      if (!lp[j]) lp[j] = i;
  }
  int n, q;
  scanf("%d%d", &n, &q);
  for (int h = 1; h <= n; h++) {
    int x;
    scanf("%d", &x);
    arr[h] = solve(x);
    int res, bit, sz = mem.size();
    for (int i = 0; i < (1 << sz); i++) {
      res = 1;
      bit = 0;
      for (int j = 0; j < sz; j++) {
        if (i >> j & 1)
          res *= mem[j];
        else
          bit++;
      }
      for (int j = 0; j < 8; j++) umax(ans[j + bit][h], last[j][res]);
      last[bit][res] = h;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 15; j++) umax(ans[j][i], ans[j][i - 1]);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int i = 0; i < 15; i++)
      if (ans[i][r] >= l) {
        printf("%d\n", i);
        break;
      }
  }
  return 0;
}