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;
int n;
int q;
int ans[200000][20];
int dv[6000000];
vector<int> p[200000];
int a[200000];
vector<pair<int, int> > z[6000000];
int seg[1000000][20];
void update(int ind, int l, int r, int x, int y, int u) {
  int i = ind;
  if (l == r) {
    seg[ind][u] = max(seg[ind][u], x);
    return;
  }
  i = (l + r) >> 1;
  if (y >= i + 1)
    update(2 * ind + 2, i + 1, r, x, y, u);
  else
    update(2 * ind + 1, l, i, x, y, u);
  seg[ind][u] = max(seg[ind][u], max(seg[2 * ind + 1][u], seg[2 * ind + 2][u]));
}
int query(int ind, int l, int r, int x, int y, int v) {
  if (l == r) {
    return seg[ind][v];
  }
  if (x <= l && y >= r) {
    return seg[ind][v];
  }
  int m = (l + r) >> 1;
  if (x >= m + 1) return query(2 * ind + 2, m + 1, r, x, y, v);
  if (y <= m) return query(2 * ind + 1, l, m, x, y, v);
  return max(query(2 * ind + 1, l, m, x, y, v),
             query(2 * ind + 2, m + 1, r, x, y, v));
}
int main() {
  scanf("%d %d", &n, &q);
  int l = 1, u, r;
  for (int i = 0; i < 1000000; i++) {
    for (int j = 0; j < 20; j++) seg[i][j] = -1;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    for (int j = 2; j * j <= a[i]; j++) {
      while (a[i] % (j * j) == 0) {
        a[i] /= (j * j);
      }
    }
    for (int j = 1; j * j <= a[i]; j++) {
      if (a[i] % j == 0) {
        p[i].push_back(j);
        if (j * j != a[i]) p[i].push_back(a[i] / j);
      }
    }
  }
  for (int i = 2; i < 6000000; i++) {
    if (dv[i] == 0) {
      for (int j = i; j < 6000000; j += i) dv[j]++;
    }
  }
  for (int j = 0; j < 20; j++) ans[0][j] = -1;
  for (int i = 0; i < n; i++) {
    if (i > 0) {
      for (int j = 0; j < 20; j++) ans[i][j] = ans[i - 1][j];
    }
    for (int j = 0; j < p[i].size(); j++) {
      for (int k = 0; k < z[p[i][j]].size(); k++) {
        l = z[p[i][j]][k].first;
        u = z[p[i][j]][k].second + dv[a[i]];
        ans[i][u] = max(ans[i][u], l);
      }
    }
    for (int j = 0; j < p[i].size(); j++) {
      while (z[p[i][j]].size() != 0 &&
             z[p[i][j]][z[p[i][j]].size() - 1].second >=
                 dv[a[i]] - 2 * dv[p[i][j]])
        z[p[i][j]].pop_back();
      z[p[i][j]].push_back(make_pair(i, dv[a[i]] - 2 * dv[p[i][j]]));
    }
  }
  int v = 20;
  int x = 0;
  int y = 19;
  int m;
  while (q--) {
    scanf("%d %d", &l, &r);
    l--;
    r--;
    for (int i = 0; i < 20; i++) {
      if (ans[r][i] >= l) {
        v = i;
        i = 20;
      }
    }
    printf("%d\n", v);
  }
  return 0;
}