Problem:
Score: 300 points

Problem Statement

M-kun is a student in Aoki High School, where a year is divided into N terms.

There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows:

For the first through (K-1)-th terms: not given.

For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term.

M-kun scored A_i in the exam at the end of the i-th term.

For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is strictly greater than the grade for the (i-1)-th term.

Constraints

2 \leq N \leq 200000

1 \leq K \leq N-1

1 \leq A_i \leq 10^{9}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 A_3 \ldots A_N

Output

Print the answer in N-K lines.

The i-th line should contain Yes if the grade for the (K+i)-th term is greater than the grade for the (K+i-1)-th term, and No otherwise.

Sample Input 1

5 3
96 98 95 100 20

Sample Output 1

Yes
No

His grade for each term is computed as follows:

3-rd term: (96 \times 98 \times 95) = 893760

4-th term: (98 \times 95 \times 100) = 931000

5-th term: (95 \times 100 \times 20) = 190000

Sample Input 2

3 2
1001 869120 1001

Sample Output 2

No

Note that the output should be No if the grade for the 3-rd term is equal to the grade for the 2-nd term.

Sample Input 3

15 7
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9

Sample Output 3

Yes
Yes
No
Yes
Yes
No
Yes
Yes

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define fio                         \
  ios_base::sync_with_stdio(false); \
  cin.tie(NULL);                    \
  cout.tie(NULL);
#define ll long long
#define ull unsigned long long
#define ld long double
#define endl "\n"
#define ff first
#define ss second
#define pi pair<ll, ll>
#define PI 3.141592653589793238462643383279502884L
#define mp make_pair
#define pb push_back
#define fall(i, a, b) for (int i = a; i < b; i++)
#define sz(x) (ll) x.size()
#define all(x) x.begin(), x.end()
#define ed cout << "ed";
#define time_passed 1.0 * clock() / CLOCKS_PER_SEC
ll MOD = 1e9 + 7;
ll MOD1 = 1e9 + 123;
ll MOD2 = 1e9 + 321;

int main()
{
  fio;
  ll t, a, n, q, k, i, l, m, c, u, f, j, p, r, x, y, s, b, d;
  cin>>n>>k;
  vector<ll>v(n,0);
  for(i=0;i<n;i++)
  cin>>v[i];c=1;
  for(i=0;i<=k-1;i++)
  c*=v[i];f=c;
  for(i=k;i<n;i++)
  {
   f/=v[i-k];
    f*=v[i];
	if(f>c)cout<<"Yes\n";
	else cout<<"No\n";
     c=f;
  }
  return 0;
}