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;
inline long long rd() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
void writeln(long long x) {
  write(x);
  puts("");
}
const int maxai = 5032999;
bool isp[maxai];
int p[maxai], a[200005], wzp[266], n, q;
int f[200005][17], res[maxai][8];
void init() {
  memset(isp, true, sizeof(isp));
  for (int i = (2); i < int(maxai); ++i) {
    if (isp[i]) {
      p[++*p] = i;
      for (int j = i + i; j < maxai; j += i) isp[j] = false;
    }
  }
  for (int i = (0); i < int(1 << 8); ++i) wzp[i] = wzp[i >> 1] + (i & 1);
}
signed main() {
  init();
  n = rd(), q = rd();
  for (int i = (1); i <= int(n); ++i) a[i] = rd();
  for (int id = (1); id <= int(n); ++id) {
    int top = 0, s[10];
    for (int j = 1; 1ll * p[j] * p[j] <= a[id] && j <= *p; ++j) {
      while (a[id] % (p[j] * p[j]) == 0) a[id] /= p[j] * p[j];
      if (a[id] % p[j] == 0) s[top++] = p[j], a[id] /= p[j];
    }
    if (a[id] != 1) s[top++] = a[id];
    for (int i = (0); i < int(1 << top); ++i) {
      int ans = 1;
      for (int j = 0; j < top; ++j)
        if (i >> j & 1) ans *= s[j];
      for (int j = (0); j < int(8); ++j)
        f[id][j + top - wzp[i]] = ((f[id][j + top - wzp[i]]) < (res[ans][j])
                                       ? (res[ans][j])
                                       : (f[id][j + top - wzp[i]]));
      res[ans][top - wzp[i]] = id;
    }
  }
  for (int i = (1); i <= int(n); ++i) {
    for (int j = (1); j < int(16); ++j)
      f[i][j] = ((f[i][j]) < (f[i][j - 1]) ? (f[i][j - 1]) : (f[i][j]));
    for (int j = (0); j < int(16); ++j)
      f[i][j] = ((f[i][j]) < (f[i - 1][j]) ? (f[i - 1][j]) : (f[i][j]));
  }
  for (int id = (0); id < int(q); ++id) {
    int l = rd(), r = rd();
    for (int i = (0); i < int(16); ++i)
      if (f[r][i] >= l) {
        writeln(i);
        break;
      }
  }
  return 0;
}