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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define MS(ARRAY,VALUE) memset(ARRAY,VALUE,sizeof(ARRAY))
#define Fin freopen("input.txt","r",stdin)
#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>


#define MOD 1000000007ll
#define MX 2000006 //200005

ll m, n;
ll a[MX];
int main()
{
    #ifndef ONLINE_JUDGE
    //Fin;
    #endif
    int test=1;
    ll x,y, k, ans;
    ll r,g,b,i;
    //cin>>test;
    //test=1;
    while(test--)
    {
        r=1;
        cin>>n>>k;
        for(i=1;i<=n;i++)scanf("%lld",&a[i]);
         for(i=1;i<=k;i++)r*=a[i];
        for(i=k+1;i<=n;i++)
        {
            g=r/(a[i-k]);
            g*=a[i];
            if(g>r)puts("Yes");
            else puts("No");
        }

    }
    return 0;
}