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>
#pragma GCC optimize - O3
using namespace std;
using ll = int64_t;
using ld = double;
using ull = uint64_t;
#pragma GCC optimize - O3
using namespace std;
using ll = int64_t;
using ld = double;
using ull = uint64_t;
const int MAXN = 5500228;
int fen[MAXN];
int p[MAXN];
void precalc() {
  for (int i = 2; i < MAXN; ++i) {
    if (p[i] == 0) {
      p[i] = i;
      for (ll j = ll(i) * i; j < MAXN; j += i) {
        if (p[j] == 0) {
          p[j] = i;
        }
      }
    }
  }
}
int n;
void upd(int x, int y) {
  for (int i = x; i < n; i |= (i + 1)) {
    fen[i] = min(fen[i], y);
  }
}
int gt(int x) {
  int ans = MAXN;
  for (int i = x + 1; --i >= 0; i &= (i + 1)) {
    ans = min(ans, fen[i]);
  }
  return ans;
}
const int MOD = 1e9 + 7;
using arr = array<int, 8>;
map<int, arr> mm;
vector<pair<int, int>> q[MAXN];
int ans[MAXN];
int a[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(20);
  precalc();
  int qc;
  cin >> n >> qc;
  fill(begin(fen), end(fen), MAXN);
  precalc();
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < qc; ++i) {
    int l, r;
    cin >> l >> r;
    q[l - 1].emplace_back(r - 1, i);
  }
  for (int i = n - 1; i >= 0; --i) {
    int c = a[i];
    vector<int> f;
    while (c != 1) {
      int pp = p[c];
      bool ad = false;
      while (pp == p[c]) {
        c /= pp;
        ad ^= 1;
      }
      if (ad) {
        f.push_back(pp);
      }
    }
    assert(f.size() < 8);
    for (int m = 0; m < (1 << f.size()); ++m) {
      int val = 1;
      int ad = f.size();
      for (int j = 0; j < f.size(); ++j) {
        if (m & (1 << j)) {
          val *= f[j];
          ad -= 2;
        }
      }
      auto& ar = mm[val];
      for (int j = 0; j < 8; ++j) {
        if (ar[j]) {
          upd(ar[j], j + ad);
        }
      }
      ar[f.size()] = i;
    }
    for (auto& [r, id] : q[i]) {
      ans[id] = gt(r);
    }
  }
  for (int i = 0; i < qc; ++i) {
    cout << ans[i] << "\n";
  }
}