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>
const int md = 1e9 + 7;
const long long inf = 4e18;
const int OO = 0;
const int OOO = 1;
using namespace std;
const int M = 5050505;
int sieve[M] = {};
void prep() {
  for (int i = 2; i * i <= M; i++) {
    if (!sieve[i]) {
      sieve[i] = i;
      for (int j = i * i; j < M; j += i) sieve[j] = i;
    }
  }
}
vector<int> factor(int v) {
  vector<int> rtn;
  int p, c;
  while (sieve[v]) {
    p = sieve[v];
    c = 0;
    while (v % p == 0) v /= p, c++;
    if (c & 1) rtn.push_back(p);
  }
  if (v > 1) rtn.push_back(v);
  return rtn;
}
void factors(const vector<int> &f, int i, int val, int extra,
             vector<pair<int, int>> &res) {
  if (i == f.size()) {
    res.emplace_back(val, extra);
    return;
  }
  factors(f, i + 1, val, extra + 1, res);
  factors(f, i + 1, val * f[i], extra, res);
}
int best[M][8];
int n, q;
vector<int> a;
int R[200005][15];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  prep();
  for (int i = 0; i < M; i++)
    for (int j = 0; j < 8; j++) best[i][j] = md;
  cin >> n >> q;
  a.resize(n);
  for (auto &i : a) cin >> i;
  for (int j = 0; j < 15; j++) R[n - 1][j] = md;
  vector<int> f = factor(a[n - 1]);
  vector<pair<int, int>> sub;
  factors(f, 0, 1, 0, sub);
  for (const auto &i : sub)
    best[i.first][i.second] = min(best[i.first][i.second], n - 1);
  for (int i = n - 2; i >= 0; i--) {
    for (int j = 0; j < 15; j++) R[i][j] = R[i + 1][j];
    f = factor(a[i]);
    sub.clear();
    factors(f, 0, 1, 0, sub);
    for (const auto &p : sub) {
      for (int j = 0; j < 8; j++)
        R[i][p.second + j] = min(R[i][p.second + j], best[p.first][j]);
    }
    for (const auto &p : sub)
      best[p.first][p.second] = min(best[p.first][p.second], i);
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    for (int i = 0; i < 15; i++)
      if (R[l][i] <= r) {
        cout << i << '\n';
        break;
      }
  }
}