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 MAXN = 200010, MAXQ = 1100010;
const int MAXVAL = 5100000;
const int MAXANS = 14, MAXS = 7;
struct Query {
  int idx, l, r;
} qry[MAXQ];
vector<int> fac[MAXN];
int a[MAXN], qans[MAXQ], n, q;
int maxdiv[MAXVAL];
void init() {
  memset(maxdiv, 0, sizeof(maxdiv));
  for (int x = 2; x < MAXVAL; ++x)
    if (!maxdiv[x])
      for (int y = x; y < MAXVAL; y += x) maxdiv[y] = x;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    int y = 1;
    fac[i].clear();
    for (int x = a[i]; x != 1;) {
      int d = maxdiv[x], t = 0;
      while (x % d == 0) x /= d, ++t;
      if (t & 1) fac[i].push_back(d), y *= d;
    }
    a[i] = y;
  }
  for (int i = 1; i <= q; ++i) {
    qry[i].idx = i;
    scanf("%d%d", &qry[i].l, &qry[i].r);
  }
  sort(qry + 1, qry + q + 1, [](Query A, Query B) { return A.r > B.r; });
}
int f[MAXANS + 1];
int lb[MAXVAL][MAXS + 1];
void dp() {
  memset(f, 0, sizeof(f));
  memset(lb, 0, sizeof(lb));
  int rest = q;
  for (int r = 1; r <= n; ++r) {
    if (!rest) break;
    int sr = fac[r].size();
    for (int mask = 0; mask < (1 << sr); ++mask) {
      int sd = __builtin_popcount(mask), d = 1;
      for (int i = 0; i < sr; ++i)
        if ((mask >> i) & 1) d *= fac[r][i];
      for (int ans = 0; ans <= MAXANS; ++ans) {
        int sl = ans + 2 * sd - sr;
        if (sl >= 0) {
          f[ans] = max(f[ans], lb[d][min(sl, MAXS)]);
          assert(d >= 0 && d < MAXVAL);
          assert(min(sl, MAXS) >= 0 && min(sl, MAXS) <= MAXS);
        }
      }
    }
    for (int mask = 0; mask < (1 << sr); ++mask) {
      int d = 1;
      for (int i = 0; i < sr; ++i)
        if ((mask >> i) & 1) d *= fac[r][i];
      for (int i = sr; i <= MAXS; ++i) lb[d][i] = r;
    }
    for (; rest > 0 && qry[rest].r == r; --rest) {
      for (int ans = 0; ans <= MAXANS; ++ans)
        if (f[ans] >= qry[rest].l) {
          qans[qry[rest].idx] = ans;
          break;
        }
    }
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", qans[i]);
}
int main() {
  init();
  dp();
  return 0;
}