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 N = 200010;
const int C = 5100000;
const int MAX_ANS = 16;
int n, q;
int primeDivisor[C];
int a[N];
void init() {
  for (int i = 2; i < C; i++) {
    if (!primeDivisor[i]) {
      for (int j = i; j < C; j += i) {
        if (!primeDivisor[j]) {
          primeDivisor[j] = i;
        }
      }
    }
  }
}
int normalize(int u) {
  int res = 1;
  while (u > 1) {
    int x = primeDivisor[u];
    int f = 0;
    while (u % x == 0) {
      u /= x;
      f ^= 1;
    }
    if (f) {
      res *= x;
    }
  }
  return res;
}
vector<int> last[C];
int dp[MAX_ANS][N];
vector<int> getDivs(int u) {
  vector<int> res;
  while (u > 1) {
    int x = primeDivisor[u];
    res.push_back(x);
    u /= x;
  }
  return res;
}
int main() {
  init();
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i] = normalize(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < MAX_ANS; j++) {
      dp[j][i] = dp[j][i - 1];
    }
    vector<int> divs = getDivs(a[i]);
    for (int mask = 0; mask < (1 << (int)divs.size()); mask++) {
      int val = 1;
      int cnt = 0;
      for (int j = 0; j < divs.size(); j++) {
        if (mask & (1 << j)) {
          val *= divs[j];
        } else {
          cnt++;
        }
      }
      for (int j = 0; j < last[val].size(); j++) {
        if (j + cnt < MAX_ANS) {
          dp[j + cnt][i] = max(dp[j + cnt][i], last[val][j]);
        }
      }
    }
    for (int mask = 0; mask < (1 << (int)divs.size()); mask++) {
      int val = 1;
      int cnt = 0;
      for (int j = 0; j < divs.size(); j++) {
        if (mask & (1 << j)) {
          val *= divs[j];
        } else {
          cnt++;
        }
      }
      if (last[val].size() <= cnt) {
        last[val].resize(cnt + 1, 0);
      }
      last[val][cnt] = i;
    }
  }
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int res = 0;
    while (dp[res][r] < l) {
      res++;
    }
    printf("%d\n", res);
  }
  return 0;
}