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 = 194598, MAXA = 5032107;
int dp[MAXA + 5][8], best[MAXN + 5][14], fac[130];
bool isP[MAXA + 5];
vector<int> p;
int n, q, x, y, l, r;
int main() {
  scanf("%d%d", &n, &q);
  memset(isP, true, sizeof isP);
  for (int i = (2); i <= (MAXA); ++i)
    if (isP[i])
      for (int j = i + i; j <= MAXA; j += i) isP[j] = false;
  for (int i = (2); i <= (MAXA); ++i)
    if (isP[i]) p.push_back(i);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &y);
    vector<int> pfac;
    for (auto x : p) {
      if (y == 1 || x * x > y) break;
      if (y % x == 0) {
        while (y % (x * x) == 0) y /= (x * x);
        if (y % x == 0) pfac.push_back(x), y /= x;
      }
    }
    if (y > 1) pfac.push_back(y);
    for (int k = (0); k <= (14); ++k) best[i][k] = best[i - 1][k];
    for (int j = (0); j <= ((1 << pfac.size()) - 1); ++j) {
      int low = 0, dist = pfac.size() - __builtin_popcount(j);
      for (; (1 << low) <= j; ++low)
        if (j >> low & 1) break;
      int val = fac[j] = (j ? fac[j - (1 << low)] * pfac[low] : 1);
      for (int k = (0); k <= (7); ++k)
        best[i][dist + k] = max(best[i][dist + k], dp[val][k]);
      dp[val][dist] = max(dp[val][dist], i);
    }
  }
  while (q--) {
    scanf("%d%d", &l, &r);
    for (int i = (0); i <= (14); ++i)
      if (best[r][i] >= l) {
        printf("%d\n", i);
        break;
      }
  }
  return 0;
}