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 = 200005, M = 5100000;
int n, m, a[N], p[M], cnt, low[M], low2[M], pc[M], f[M][9], g[17], d[N], dn,
    tmp[20], ans[M];
bool is[M];
vector<pair<int, int> > V[N];
inline void read(int &i) {
  i = 0;
  char c = getchar();
  bool j = 0;
  for (; !isdigit(c); c = getchar())
    if (c == '-') j = 1;
  for (; isdigit(c); c = getchar()) i = (i << 1) + (i << 3) + c - '0';
  i = j ? -i : i;
}
inline void up(int &x, int y) { x < y ? x = y : 0; }
inline void init() {
  int n = 5032107;
  for (int i = 2, t; i <= n; ++i) {
    if (!is[i]) p[++cnt] = i, low[i] = i, pc[i] = 1;
    for (int j = 1; j <= cnt; ++j) {
      t = i * p[j];
      if (t > n) break;
      is[t] = 1;
      low[t] = p[j];
      if (i % p[j])
        low2[t] = low2[i], pc[t] = pc[i] + 1;
      else {
        low2[t] = p[j] * p[j];
        break;
      }
    }
  }
}
inline void sim(int &x) {
  while (low2[x]) x /= low2[x];
}
inline void divisor(int x) {
  dn = 0;
  int t = 0;
  while (x > 1) tmp[++t] = low[x], x /= low[x];
  for (int i = 0; i < (1 << t); ++i) {
    int mul = 1;
    for (int j = 0; j < t; ++j)
      if (i >> j & 1) mul *= tmp[j + 1];
    d[++dn] = mul;
  }
}
int main() {
  init();
  read(n);
  read(m);
  for (int i = 1; i <= n; ++i) read(a[i]), sim(a[i]);
  for (int i = 1, l, r; i <= m; ++i)
    read(l), read(r), V[r].push_back(make_pair(l, i));
  for (int i = 1; i <= n; ++i) {
    divisor(a[i]);
    for (int j = 1; j <= dn; ++j)
      for (int k = 0; k <= 8; ++k) up(g[k + pc[a[i] / d[j]]], f[d[j]][k]);
    for (int j = 1; j <= dn; ++j) up(f[d[j]][pc[a[i] / d[j]]], i);
    for (pair<int, int> p : V[i])
      for (int j = 0;; ++j)
        if (g[j] >= p.first) {
          ans[p.second] = j;
          break;
        }
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}