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 nmax = 194598;
const int qmax = 1049658;
const int valmax = 5032107;
int n;
bool ciur[valmax + 1];
vector<int> prime;
vector<int> v[nmax + 1];
struct str {
  int val, lg;
  str() {}
  str(int _val, int _lg) { val = _val, lg = _lg; }
  bool operator==(const str &shp) const {
    return val == shp.val && lg == shp.lg;
  }
};
struct Hsh {
  size_t operator()(const str &shp) const { return shp.val + shp.lg; }
};
unordered_map<str, int, Hsh> mp;
int per[nmax + 1][15], sp[nmax + 1][15];
void precalc() {
  for (int i = 2; i * i * i * i <= valmax; ++i) {
    if (ciur[i] == 0) {
      for (int j = i * i; j * j <= valmax; j += i) ciur[j] = 1;
    }
  }
  for (int i = 2; i * i <= valmax; ++i)
    if (ciur[i] == 0) prime.push_back(i);
}
int main() {
  int m;
  fscanf(stdin, "%d%d", &n, &m);
  precalc();
  for (int i = 1; i <= n; ++i) {
    int x;
    fscanf(stdin, "%d", &x);
    for (int j = 0; j < (int)prime.size() && prime[j] * prime[j] <= x; ++j) {
      if (x % prime[j] != 0) continue;
      int e = 0;
      while (x % prime[j] == 0) {
        x /= prime[j];
        ++e;
      }
      if (e % 2) v[i].push_back(prime[j]);
    }
    if (x > 1) v[i].push_back(x);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (1 << (int)v[i].size()); ++j) {
      int prod = 1, cnt = 0;
      for (int k = 0; k < (int)v[i].size(); ++k) {
        if (j & (1 << k)) {
          prod *= v[i][k];
          ++cnt;
        }
      }
      for (int k = 0; k <= 7; ++k) {
        if (mp.find(str(prod, k)) != mp.end())
          per[i][v[i].size() + k - 2 * cnt] =
              max(per[i][v[i].size() + k - 2 * cnt], mp[str(prod, k)]);
      }
      mp[str(prod, v[i].size())] = i;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 15; ++j) sp[i][j] = max(sp[i - 1][j], per[i][j]);
  }
  for (int i = 1; i <= m; ++i) {
    int x, y;
    fscanf(stdin, "%d%d", &x, &y);
    for (int j = 0; j < 15; ++j) {
      if (sp[y][j] >= x) {
        fprintf(stdout, "%d\n", j);
        break;
      }
    }
  }
  return 0;
}