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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define test int t;cin>>t;while(t--)
#define REP(i,n) for (int i = 1; i <= n; i++)
#define MOD 1000000007
#define ll long long
#define ld long double
#define int ll
#define pb push_back
#define ii pair<int,int>
#define llll pair<ll,ll>
#define vi vector<int>
#define vl vector<ll>
#define vii vector<ii>
#define vllll vector<llll>
#define ff first
#define ss second
#define INF 1000000000
#define HINF 1000000000000000
#define mem(a,b) memset(a,b,sizeof(a))
#define arrin(x,n) int x[n]; for(int o=0;o<n;o++) cin>>x[o]
#define arrout(x,n) for(int o=0;o<n;o++) cout<<x[o]<<" "; cout<<endl
#define all(x) x.begin(),x.end()
#define deb(x) cout<<#x<<'='<<x<<endl
#define deb2(x,y) cout<<#x<<'='<<x<<" "<<#y<<'='<<y<<endl
#define deb3(x,y,z) cout<<#x<<'='<<x<<" "<<#y<<'='<<y<<" "<<#z<<'='<<z<<endl
#define debarr(arr,n) for(int o=0;o<n;o++) deb2(o,arr[o])
const double pi = 3.14159265358979323846;

ll power(ll a, ll b){//a^b
    ll res=1;
    //a=a%MOD;
    while(b>0){
        if(b&1)
		{res=(res*a);
		//res%=MOD;
		b--;}
        a=(a*a);
        //a%=MOD;
        b>>=1;
    }
    return res;
}
ll fermat_inv(ll y){return power(y,MOD-2);}
ll gcd(ll a, ll b){return (b==0)?a:gcd(b,a%b);}
int C(int n,int k)
{
	int ans=1;
	if(k>n-k)
	k=n-k;
	for(int i=1;i<=k;i++)
	ans*=(n-i+1),ans/=i;
	return ans;
}

void solve()
{

	int i=0,j=0,k=0,m,n,p,sum=0,cnt=0;
	cin>>n>>k;
	arrin(arr,n);
	for(i=k;i<n;i++)
	{
	    if(arr[i]>arr[i-k])
	    cout<<"Yes\n";
	    else
	    cout<<"No\n";
	}
	
}
int32_t main() {
    fast
    //freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w+", stdout);
	//cout<<fixed<<setprecision(6);
    //test
    {
        
        solve();
		
		      
    }
    return 0;
}