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 = 222222;
const int W = 5050000;
int m, pr[N], k, e[N], g[333], h[333];
int n, q, a[N], ans[W];
int b[W][8], c[22], d[W];
vector<pair<int, int> > v[N];
void modify(int x, int y) {
  for (int i = 0; i < 8; i = i + 1)
    if (c[k + i - d[x]] < b[x][i]) c[k + i - d[x]] = b[x][i];
  if (b[x][k] < y) b[x][k] = y;
}
int query(int x) {
  for (int i = 0; i < 22; i = i + 1)
    if (c[i] >= x) return i;
  return -1;
}
int main() {
  int i, j, x, y;
  m = 0;
  for (i = 2; i <= 3000; i = i + 1) {
    x = 1;
    for (j = 2; j * j <= i; j = j + 1)
      if (i % j == 0) x = 0;
    if (x) pr[++m] = i;
  }
  for (i = 1; i < (1 << 8); i = i + 1) g[i] = g[i ^ (i & -i)] + 2;
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; i = i + 1) scanf("%d", a + i);
  for (i = 1; i <= q; i = i + 1) {
    scanf("%d%d", &x, &y);
    v[y].push_back(make_pair(x, i));
  }
  for (i = 1; i <= n; i = i + 1) {
    k = 0;
    for (j = 1; j <= m; j = j + 1) {
      x = 0;
      while (a[i] % pr[j] == 0) {
        x++;
        a[i] /= pr[j];
      }
      if (x & 1) e[k++] = pr[j];
    }
    if (a[i] > 1) e[k++] = a[i];
    h[0] = 1;
    for (j = 0; j < k; j = j + 1) h[1 << j] = e[j];
    for (j = 1; j < (1 << k); j = j + 1) h[j] = h[j ^ (j & -j)] * h[j & -j];
    for (j = 0; j < (1 << k); j = j + 1) d[h[j]] = g[j], modify(h[j], i);
    for (j = v[i].size(); j--;) ans[v[i][j].second] = query(v[i][j].first);
  }
  for (i = 1; i <= q; i = i + 1) printf("%d\n", ans[i]);
  return 0;
}