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;
using namespace rel_ops;
using ll = int64_t;
using Pii = pair<int, int>;
using ull = uint64_t;
using Vi = vector<int>;
void run();
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  run();
  return 0;
}
int uplg(int n) { return 32 - __builtin_clz(n); }
int uplg(ll n) { return 64 - __builtin_clzll(n); }
constexpr int MAX_N = 200000;
constexpr int MAX_VAL = 6000000;
constexpr int MAX_FACTORS = 8;
int dp[MAX_VAL][MAX_FACTORS];
int prefs[MAX_FACTORS * 2];
Vi factor(MAX_VAL + 1);
void sieve() {
  for (int i = 2; i * i <= MAX_VAL; i++)
    if (!factor[i])
      for (int j = i * i; j <= MAX_VAL; j += i)
        if (!factor[j]) factor[j] = i;
  for (int i = (0); i < (MAX_VAL + 1); i++)
    if (!factor[i]) factor[i] = i;
}
Vi factorize(ll n) {
  Vi ret;
  while (n > 1) {
    int f = factor[n];
    if (ret.empty() || ret.back() != f)
      ret.push_back(f);
    else
      ret.pop_back();
    n /= f;
  }
  return ret;
}
void run() {
  int n, m;
  cin >> n >> m;
  Vi elems(n);
  for (auto& e : (elems)) cin >> e;
  vector<Pii> queries(m);
  for (auto& q : (queries)) {
    cin >> q.first >> q.second;
    q.first--;
  }
  Vi answers(m, -1);
  Vi sorted(m);
  iota((sorted).begin(), (sorted).end(), 0);
  sort((sorted).begin(), (sorted).end(),
       [&](int l, int r) { return queries[l].first > queries[r].first; });
  for (int i = (0); i < (MAX_VAL); i++)
    for (int j = (0); j < (MAX_FACTORS); j++) dp[i][j] = 1e9;
  for (int j = (0); j < (MAX_FACTORS * 2); j++) prefs[j] = 1e9;
  int cur = 0;
  sieve();
  for (int begin = n - 1; begin >= 0; begin--) {
    Vi factors = factorize(elems[begin]);
    for (int i = (0); i < (1 << int((factors).size())); i++) {
      int div = 1;
      int inter = __builtin_popcount(i);
      for (int j = (0); j < (int((factors).size())); j++)
        if ((i >> j) & 1) div *= factors[j];
      for (int count = (0); count < (MAX_FACTORS); count++) {
        int pos = dp[div][count];
        int val = int((factors).size()) + count - 2 * inter;
        prefs[val] = min(prefs[val], pos);
      }
      dp[div][int((factors).size())] = begin;
    }
    while (cur < m && queries[sorted[cur]].first == begin) {
      for (int i = (0); i < (MAX_FACTORS * 2); i++) {
        if (prefs[i] < queries[sorted[cur]].second) {
          answers[sorted[cur]] = i;
          break;
        }
      }
      cur++;
    }
  }
  for (auto& a : (answers)) cout << a << '\n';
}