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)

/*
                        ********************
			**  MANAN DIYORA  **
			**    DA-IICT     **
			**   diyora13     **
			********************
*/
#include <bits/stdc++.h>
#include <time.h>
using namespace std;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define inarr(a,n) for(int i=0;i<n;i++) cin>>a[i]
#define outarr(a,n) for(int i=0;i<n;i++) cout<<a[i]<<" "; cout<<endl
#define maxa(a,n) *max_element(a,a+n)
#define mina(a,n) *min_element(a,a+n)
#define Max(x,y,z) max(x,max(y,z))
#define Min(x,y,z) min(x,min(y,z))
#define fr(i,s,e) for(ll i=s;i<e;i++)
#define rf(i,s,e) for(ll i=s-1;i>=e;i--)
#define gcd(a,b) __gcd(a,b)
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define F first
#define S second
#define sz size()
#define len length()
#define all(v) v.begin(),v.end()

typedef long long int ll;
typedef long double ld;
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef vector<pair<long long,long long> > vpll;
// freopen("input.txt","r",stdin);
// freopen("output.txt","w",stdout);
// bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) { return (a.second < b.second); }
#define PI 3.141592653589793
#define MOD 1000000007

bool flag=0,flag1=0,flag2=0;

int main()
{
	IOS
	ll t=1;
//		cin>>t;
	while(t--)
	{
		ll n,k; cin>>n>>k;
		ll a[n]; inarr(a,n);
		ll mul=1;
		fr(i,0,k) mul*=a[i];
		fr(i,k,n)
		{
			if(mul<(mul*a[i])/a[i-k]) cout<<"Yes"<<endl;
			else cout<<"No"<<endl;
			mul*=a[i];
			mul/=a[i-k];
		}
	}
}