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)

// I make this just for fun because i'm done
// Aimi Haraguni >> Konomi Suzuki >> Yui >>  Ikimono Gakari >> Garnidelia >> Kalafina >> Eir Aoi.   .. dude? 
// problems that involves any kind of persistent data structures are the best of the best, are not them?
 
// #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
 
#define pb push_back
#define ff  first
#define ss second
#define tm1 first
#define tm2 second.first
#define tm3 second.second
#define sz(x) ll(x.size())
#define all(v) (v).begin(), (v).end()
// #define fill(x, v) memset(x, v, sizeof(x))
#define FER(i,a,b) for(int i=int(a); i< int(b); ++i)
#define IFR(i,a,b) for(int i=int(a); i>=int(b); --i )
#define fastio ios_base::sync_with_stdio(0); cin.tie(0)
 
// #define N 21000000
// #define M 19
// #define mod 
#define mod  998244353
#define mod1 1000000007
#define mod2 1000000009
#define bas 987625403
#define sqr(x) (x)*(x)  
#define INF 3000000000000000000
 
using namespace std;
using namespace __gnu_pbds;
 
typedef int ll;
typedef pair<ll, ll> ii;
typedef pair<ll, ii > tri;
typedef vector<ll> vi;
typedef vector<vi> vv;
typedef vector<ii> vii;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> S_t;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
 
#define trace(...) fff(#__VA_ARGS__, __VA_ARGS__)
template<typename t> void fff(const char* x, t&& val1) { cout<<x<< " : "<<val1<<"\n";}
template<typename t1, typename... t2> void fff(const char* x, t1&& val1, t2&&... val2){
    const char* xd=strchr(x+1, ',');
    cout.write(x, xd-x)<<" : "<<val1<<" | ";
    fff(xd+1, val2...);
}
 
inline ll add(ll a, ll b, ll m) { return a+b < m? a+b : a+b-m;}
inline ll rem(ll a, ll b, ll m) { return a >= b? a - b: a- b + m;}
inline ll mul(ll a, ll b, ll m) { return (long long) a*b%m;}
inline void Mul(ll &a, ll b, ll m) { a = (long long) a*b%m;}
inline ll bp(ll a, ll p, ll m){
	ll ret;
	for(ret = 1; p; p>>=1, Mul(a, a, m)) (p & 1) && (Mul(ret, a, m), 1);
	return ret;
}
 
static inline void read(ll &x) { cin>>x;}

long double dp[1<<19];
inline long double Que(ll l, ll r){
	if(l) return dp[r-1] - dp[l-1];
	return dp[r-1];
}
int main(){
	fastio;
	ll n, k; read(n), read(k);
	long double ans = 0;
	FER(i, 0, n){
		ll x; read(x);
		ans += log2(x);
		dp[i] = ans;
	}
	FER(i, 0, n-k){
		long double a, b;
		a = Que(i, i+k);
		b = Que(i+1, i+1+k);
		if(a + 1e-10 < b){
			cout<<"Yes\n";
		}
		else{
			cout<<"No\n";
		}
	}
	return 0;
}