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;

typedef long long int li;
typedef long double ld;
typedef vector<li> vi;
typedef pair<li,li> pi;
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define B begin()
#define E end()
#define f(i,l,r) for(li i=l;i<=r;++i)
#define fr(i,l,r) for(li i=l;i>=r;--i)
#define u_map unordered_map
#define endl "\n"
#define debug(x) cout<<#x<<" = "<<x<<endl;

li fastpow(li base,li exp,li M) 
{
    li res=1;
    while(exp>0) {
        if(exp&1)
            res=(res*base)%M;

        base=(base*base)%M;
        exp>>=1;
    }
    return res;
}

void solve()
{
    li n,k;

    cin>>n>>k;

    li a[n];

    f(i,0,n-1)
    {
    	cin>>a[i];
    }

    f(i,k,n-1)
    {
    	if(a[i]>a[i-k])
    		cout<<"Yes\n";
    	else
    		cout<<"No\n";
    }
}

int main()
{    
   

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    li t=1;

    // cin>>t;

    while(t--)
    {
        solve();
    }

    return 0;
}