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>
#define Imposter ios_base::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long int
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("O3")
using namespace __gnu_pbds;
#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>//s.order_of_key(val) *s.find_by_order(ind)
using namespace std;
#define f(n) for(ll i=0;i<n;i++)
#define rep(i,a,b) for(i=a;i<b;i++)
#define input(arr,n) for(ll i1=0;i1<n;i1++ )cin>>arr[i1]
#define vll vector<ll>
#define vlll vector<vll>
#define all(x) x.begin(),x.end()
ll power(ll x, ll y, ll p) { 
    ll res = 1;    x = x % p;  
    while(y > 0) {
        if(y & 1) res = (res*x) % p; 
        y = y>>1;    x = (x*x) % p; 
    } return res; 
}
int main()
{
    Imposter
    
    int t=1;
    //cin>>t;
    while(t--)
    {
       ll n,k;
       cin>>n>>k;
       ll a[n];
       input(a,n);
       deque<ll>ans;
       f(k)
       {
           ans.push_back(a[i]);
       }
       for(int i=k;i<n;i++)
       {
           if(ans.front()>=a[i])
           cout<<"No"<<endl;
           else
           cout<<"Yes"<<endl;
           ans.pop_front();
           ans.push_back(a[i]);
       }
       
        
    }
    return 0;
}