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 mod = 998244353;
const int inf = 1 << 30;
const int maxn = 200000 + 5;
const int maxv = 5032107 + 5;
namespace Factor {
const int maxp = 5032107 + 5;
int vis[maxp], prime[maxp], pcnt;
int mdiv[maxp], ndiv[maxp];
void init() {
  for (int i = 2; i < maxp; i++) {
    if (!vis[i]) prime[pcnt++] = i, mdiv[i] = i;
    for (int j = 0; j < pcnt && prime[j] * i < maxp; j++) {
      vis[i * prime[j]] = 1;
      mdiv[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 2; i < maxp; i++) {
    int p = mdiv[i];
    int cur = i / mdiv[i];
    if (cur % p == 0)
      ndiv[i] = ndiv[cur];
    else
      ndiv[i] = ndiv[cur] + 1;
  }
}
int norm(int x) {
  int tot = 1;
  for (int i = 0; i < pcnt && 1ll * prime[i] * prime[i] <= x; i++) {
    if (x % prime[i]) continue;
    int c = 0;
    while (x % prime[i] == 0) c++, x /= prime[i];
    if (c % 2) tot *= prime[i];
  }
  if (x > 1) tot *= x;
  return tot;
}
vector<int> getFactor(int x) {
  int tot = x;
  vector<int> v;
  for (int i = 1; 1ll * i * i <= tot; i++) {
    if (tot % i) continue;
    v.push_back(i);
    if (1ll * i * i != tot) v.push_back(tot / i);
  }
  return v;
}
}  // namespace Factor
using Factor::ndiv;
int n, q, a[maxn], ans[1049659];
int dp[maxv][8], best[30];
vector<pair<int, int> > que[maxn];
int main() {
  Factor::init();
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1, l, r; i <= q; i++) {
    scanf("%d%d", &l, &r);
    que[r].emplace_back(l, i);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = Factor::norm(a[i]);
    auto divs = Factor::getFactor(a[i]);
    for (auto& d : divs) {
      int tot = ndiv[a[i]] - ndiv[d];
      for (int i = 0; i <= 7; i++) {
        best[i + tot] = max(best[i + tot], dp[d][i]);
      }
    }
    for (auto& q : que[i]) {
      int le = q.first, id = q.second;
      ans[id] = -1;
      for (int i = 0; i <= 14; i++) {
        if (best[i] >= le) {
          ans[id] = i;
          break;
        }
      }
      assert(ans[id] != -1);
    }
    for (auto& d : divs) {
      int ad = ndiv[a[i]] - ndiv[d];
      dp[d][ad] = max(dp[d][ad], i);
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}