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 = 2e5 + 5;
const int M = 1.5e6 + 5;
const unsigned long long Add1 = 998244353;
const unsigned long long Mod1 = 1919810111;
const unsigned long long Add2 = 19260817;
const unsigned long long Mod2 = 1e9 + 7;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10)
    putchar(x + '0');
  else {
    write(x / 10);
    putchar(x % 10 + '0');
  }
}
void print(int x, char c = '\n') {
  write(x);
  putchar(c);
}
int Min[20];
int ans[M];
vector<pair<int, int>> q[N];
unordered_map<unsigned long long, int> mp[8];
vector<int> a[N];
signed main() {
  memset(Min, 0x3f, sizeof(Min));
  int n = read(), Q = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    for (int j = 2; j * j <= x; j++)
      if (x % j == 0) {
        int cnt = 0;
        while (x % j == 0) x /= j, cnt++;
        if (cnt & 1) a[i].push_back(j);
      }
    if (x > 1) a[i].push_back(x);
    sort(a[i].begin(), a[i].end());
  }
  for (int i = 1; i <= Q; i++) {
    int l = read(), r = read();
    q[l].push_back(make_pair(r, i));
  }
  for (int i = n; i >= 1; i--) {
    int siz = a[i].size();
    for (int j = 0; j < (1 << siz); j++) {
      int cnt = 0;
      unsigned long long hash1 = 0, hash2 = 0;
      for (int k = 0; k < siz; k++)
        if ((j >> k) & 1)
          hash1 = (hash1 * 5032108LLu + a[i][k] + Add1) % Mod1, cnt++;
      for (int k = 0; k < siz; k++)
        if ((j >> k) & 1)
          hash2 = (hash2 * 5032108LLu + a[i][k] + Add2) % Mod2, cnt++;
      unsigned long long hash = hash1 * Mod2 + hash2;
      for (int k = 0; k <= 7; k++)
        if (mp[k].count(hash)) {
          int tot = siz + k - cnt;
          Min[tot] = min(Min[tot], mp[k][hash]);
        }
      mp[siz][hash] = i;
    }
    for (auto j : q[i])
      for (int k = 0; k <= 14; k++)
        if (Min[k] <= j.first) {
          ans[j.second] = k;
          break;
        }
  }
  for (int i = 1; i <= Q; i++) print(ans[i]);
  return 0;
}