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;
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return a < b ? b : a;
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return a < b ? a : b;
}
const char lf = '\n';
namespace ae86 {
const int bufl = 1 << 15;
char buf[bufl], *s = buf, *t = buf;
inline int fetch() {
  if (s == t) {
    t = (s = buf) + fread(buf, 1, bufl, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
inline int ty() {
  int a = 0;
  int b = 1, c = fetch();
  while (!isdigit(c)) b ^= c == '-', c = fetch();
  while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
  return b ? a : -a;
}
}  // namespace ae86
using ae86::ty;
const int _ = 199999, __ = 1049999, V = 5033333, P = 12;
int n, qn, ql[__], qr[__], ans[__] = {0}, val[_], f[V][P] = {0},
                           xloc[P + P] = {0};
vector<int> qry[__], zsy[_];
int main() {
  ios::sync_with_stdio(0), cout.tie(nullptr);
  n = ty(), qn = ty();
  for (int i = 1; i <= n; i++) val[i] = ty();
  for (int i = 1; i <= qn; i++)
    ql[i] = ty(), qr[i] = ty(), qry[qr[i]].emplace_back(i);
  for (int i = 1; i <= n; i++) {
    int x = val[i];
    for (int j = 2; 1ll * j * j <= x; j++) {
      if (x % j) continue;
      int cnts = 0;
      while (x % j == 0) x /= j, cnts++;
      if (cnts & 1) zsy[i].emplace_back(j);
    }
    if (x > 1) zsy[i].emplace_back(x);
  }
  for (int i = 1; i <= n; i++) {
    int m = zsy[i].size(), mm = 1 << m;
    for (int j = 0; j < mm; j++) {
      int a = 1, b = 0;
      for (int k = 0; k < m; k++) {
        if ((((j) >> (k)) & 1))
          a *= zsy[i][k];
        else
          b++;
      }
      for (int k = 0; k < P; k++) xloc[k + b] = max(xloc[k + b], f[a][k]);
    }
    for (auto j : qry[i]) {
      int now = 0;
      while (xloc[now] < ql[j]) now++;
      ans[j] = now;
    }
    for (int j = 0; j < mm; j++) {
      int a = 1, b = 0;
      for (int k = 0; k < m; k++) {
        if ((((j) >> (k)) & 1))
          a *= zsy[i][k];
        else
          b++;
      }
      f[a][b] = i;
    }
  }
  for (int i = 1; i <= qn; i++) cout << ans[i] << lf;
  return 0;
}