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 MAX = 5100000;
const int K = 7;
const int R = 2 * K + 1;
const int MAXN = 200000;
const int MAXS = 1 << K;
int c[MAX];
int f[MAXN][R];
int g[K + 1][MAX];
int hb[MAXS];
int cnt[MAXS];
vector<int> RMQ(const vector<int>& array,
                const vector<pair<int, int>>& queries) {
  int n = array.size();
  vector<int> dad(n);
  function<int(int)> findset = [&](int x) {
    return x == dad[x] ? x : dad[x] = findset(dad[x]);
  };
  vector<vector<int>> qs(n);
  for (int i = (0); i < ((int)queries.size()); ++i)
    qs[queries[i].second].push_back(i);
  vector<int> ans(queries.size(), -1);
  vector<int> stack;
  for (int i = (0); i < (n); ++i) {
    dad[i] = i;
    while (stack.size() && array[stack.back()] > array[i]) {
      dad[stack.back()] = i;
      stack.pop_back();
    }
    stack.push_back(i);
    for (int qi : qs[i]) ans[qi] = array[findset(queries[qi].first)];
  }
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  for (int i = (1); i < (MAXS); ++i) {
    int j = 0;
    while (((i >> j) & 1) == 0) j++;
    hb[i] = j;
    cnt[i] = cnt[i ^ (1 << j)] + 1;
  }
  c[1] = 1;
  for (int i = 2; i * i < MAX; ++i) {
    if (!c[i]) {
      for (int j = i * i; j < MAX; j += i) {
        c[j] = i;
      }
    }
  }
  int N, Q;
  cin >> N >> Q;
  memset(g, -1, sizeof(g));
  for (int i = (0); i < (N); ++i) {
    int x;
    cin >> x;
    vector<int> v;
    while (x > 1) {
      int f = c[x] == 0 ? x : c[x];
      int cnt = 0;
      while (x % f == 0) {
        x /= f;
        cnt++;
      }
      if (cnt % 2) {
        v.push_back(f);
      }
    }
    int m = v.size();
    assert(m <= 7);
    for (int j = (0); j < (R); ++j) f[i][j] = -1;
    static int u[MAXS];
    u[0] = 1;
    for (int s = (0); s < (1 << m); ++s) {
      if (s == 0) {
        u[s] = 1;
      } else {
        u[s] = u[s ^ (1 << hb[s])] * v[hb[s]];
      }
      int c = m - cnt[s];
      for (int k = (0); k < (K + 1); ++k) {
        f[i][k + c] = max(f[i][k + c], g[k][u[s]]);
      }
      g[c][u[s]] = i;
    }
  }
  vector<pair<int, int>> q(Q);
  vector<int> ans(Q, R);
  for (int i = (0); i < (Q); ++i) {
    cin >> q[i].first >> q[i].second;
    --q[i].first, --q[i].second;
  }
  for (int i = (0); i < (R); ++i) {
    vector<int> a(N);
    for (int j = (0); j < (N); ++j) a[j] = N - f[j][i];
    vector<int> ret(move(RMQ(a, q)));
    for (int j = (0); j < (Q); ++j) {
      if (N - ret[j] >= q[j].first) {
        ans[j] = min(ans[j], i);
      }
    }
  }
  for (int j = (0); j < (Q); ++j) {
    cout << ans[j] << "\n";
  }
  return 0;
}