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<iostream>
#include<string>
#include<algorithm>    
#include<cmath>
#include<map>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<stack>
#include<queue>
#include<tuple>
#include<cassert>
#include<set>
#include<functional>
//#include<bits/stdc++.h>
#define rep(i, x) for(ll i = 0; i < x; i++)
#define rep2(i, x) for(ll i = 1; i <= x; i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1000000000000000000;
const ll mod = 1000000007;
const ld pi = 3.141592653589793238;

/*bool IsPrime(ll num) {
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く
	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2) {
		if (num % i == 0) { // 素数ではない
			return false;
		}
	}
	// 素数である
	return true;
}*/
/*ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (a % b == 0)return b;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}*/
/*ll modpow(ll x, ll y) {
	ll res = 1;
	while (y) {
		if (y % 2) { res *= x; res %= mod; }
		x = x * x % mod; y /= 2;
	}
	return res;
}*/
/*ll keta(ll n) {
	ll res = 0;
	while (n >= 1) {
		res += n % 10; n /= 10;
	}
	return res;
}*/
/*ll nCk(ll n, ll k) {
	ll a = 1, b = 1;
	for (int h = n - k + 1; h <= n; h++) { a *= h; a %= mod; }
	for (int h = 1; h <= k; h++) { b *= h; b %= mod; }
	return a * modpow(b, mod - 2) % mod;
}*/
//printf("%.10f\n", n);

ll test[12345678];
signed main() {
	string s, t;
	ll n, m, k, a = 0, b = 0, c = 0, cnt = 0;
	cin >> n >> k;
	rep(i, n) { cin >> test[i]; }
	for (int h = k; h < n; h++) {
		if (test[h - k] < test[h]) { cout << "Yes" << endl; }
		else { cout << "No" << endl; }
	}
	return 0;
}