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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <chrono>

using namespace __gnu_pbds;
using namespace std;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the kth largest element.(0-based)

#define ll long long int
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define pi pair<ll,ll>
#define pii pair<ll,pi >
#define ppi pair<pi,ll>
#define ppp pair<pi,pi >
#define ex(str,a,b) str.substr((int)(a),min((int)str.size()-a,(int)(b-a+1))) 
#define all(X) X.begin(),X.end()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

// for pair comparison function(ascending order) use return (i1.ff < i2.ff);

/* string operations :
   str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
   __builtin_popcount(n) : no. of set bits in n.
*/

const int M = (1<<20)+5;
const int md = 1e9+7;

priority_queue<ll,vector<ll >,greater<ll > > pq;

ll pwr(ll a,ll n,ll m)
{
	ll p=1;
	while(n>0)
	{
		if(n%2==1)
			p=(p*a)%m;
		a=(a*a)%m;
		n=n/2;
	}
	return p;
}

ll ar[M];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	ll n,k,i;
	cin >> n >> k;
	for(i=1;i<=n;++i)
	{
		cin >> ar[i];
		if(i>k)
		{
			if(ar[i]>ar[i-k])
				cout << "Yes" << "\n";
			else
				cout << "No" << "\n";
		}
	}
}