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, M = 5100000;
template <typename T>
bool cmax(T &a, T b) {
  return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool cmin(T &a, T b) {
  return (a > b) ? a = b, 1 : 0;
}
template <typename T>
T read() {
  T ans = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch))
    ans = (ans << 3) + (ans << 1) + (ch - '0'), ch = getchar();
  return ans * f;
}
template <typename T>
void write(T x, char y) {
  if (x == 0) {
    putchar('0'), putchar(y);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  static char wr[20];
  int top = 0;
  for (; x; x /= 10) wr[++top] = x % 10 + '0';
  while (top) putchar(wr[top--]);
  putchar(y);
}
void file() {}
int n, q;
int a[N];
void input() {
  n = read<int>(), q = read<int>();
  for (register int i = (int)(1); i <= (int)(n); ++i) a[i] = read<int>();
}
int p[M + 5], ans[N][13], pos[M + 5][13], L[13];
vector<int> E;
void dfs(int dep, int id, int mul, int cnt) {
  if (dep == (int)E.size()) {
    for (int i = 0; i + cnt < 13; i++) cmax(L[i + cnt], pos[mul][i]);
    pos[mul][cnt] = id;
  } else {
    dfs(dep + 1, id, mul * E[dep], cnt);
    dfs(dep + 1, id, mul, cnt + 1);
  }
}
void work() {
  for (register int i = (int)(2); i <= (int)(M); ++i)
    if (!p[i])
      for (int j = i + i; j <= M; j += i) p[j] = i;
  int x, y, c;
  for (register int i = (int)(1); i <= (int)(n); ++i) {
    x = a[i];
    E.clear();
    while (p[x]) {
      y = p[x];
      c = 0;
      while (x % y == 0) x /= y, c ^= 1;
      if (c) E.push_back(y);
    }
    if (x > 1) E.push_back(x);
    dfs(0, i, 1, 0);
    for (register int j = (int)(0); j <= (int)(12); ++j) ans[i][j] = L[j];
  }
  int l, r, now;
  while (q--) {
    l = read<int>(), r = read<int>();
    for (now = 0; ans[r][now] < l; now++)
      ;
    write(now, '\n');
  }
}
int main() {
  file();
  input();
  work();
  return 0;
}