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>
using namespace std;
using namespace __gnu_pbds;

// Optimization

//#pragma GCC optimize("O3")
#define endl '\n'

// Shortcut

#define int long long
#define eb emplace_back
#define pb push_back
#define pob pop_back
#define mp make_pair
#define upb upper_bound
#define lwb lower_bound
#define fi first
#define se second
#define For(i, l, r) for (int i = l; i < r; i++)
#define ForE(i, l, r) for (int i = l; i <= r; i++)
#define Ford(i, r, l) for (int i = r; i > l; i--)
#define FordE(i, r, l) for (int i = r; i >= l; i--)
#define Fora(i, a) for (auto i : a)

// I/O & Debug

#define PrintV(a) Fora(iiii, a) cout << iiii << ' '; cout << endl;
#define PrintVl(a) Fora(iiii, a) cout << iiii << endl;
#define PrintA(a, l, r) for (int iiii = l; iiii <= r; iiii++) cout << a[iiii] << ' '; cout << endl;
#define PrintAl(a, l, r) for (int iiii = l; iiii <= r; iiii++) cout << a[iiii] << endl;
#define Ptest(x) return cout << x, 0;
#define gl(s) getline(cin, s);
#define setpre(x) fixed << setprecision(x)
/*
#define debug(args...){ string _sDEB = #args; replace(_sDEB.begin(), _sDEB.end(), ',', ' '); stringstream _ssDEB(_sDEB); istream_iterator<string> _itDEB(_ssDEB); DEB(_itDEB, args); }
void DEB(istream_iterator<string> it) {}
template<typename T, typename... Args>
void DEB(istream_iterator<string> it, T a, Args... args){
    cout << *it << " = " << a << endl;
    DEB(++it, args...);
}
*/

// Functions

//#define isvowel(a) (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u')
#define bend(a) a.begin(), a.end()
#define rbend(a) a.rbegin(), a.rend()
#define mset(a) memset(a, 0, sizeof(a))
#define mset1(a) memset(a, 1, sizeof(a))
#define msetn1(a) memset(a, -1, sizeof(a))
#define msetinf(a) memset(a, 0x3f, sizeof(a))
#define gcd __gcd
#define __builtin_popcount __builtin_popcountll
//mt19937 rando(chrono::steady_clock::now().time_since_epoch().count());
//int randt(int l, int r){ return (rando() % (r - l + 1) + l); }

// Data Structure

#define pque priority_queue
#define mts multiset
#define y0 _y0_
#define y1 _y1_
#define div divi
typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <ld> vld;
typedef vector <string> vs;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef vector <vi > vvi;
typedef vector <vll > vvll;
typedef vector <pii > vpii;
typedef vector <pll > vpll;

const int N = 2e5 + 5;
int mod = 1e9 + 7, mod1 = 998244353, mod2 = 1e9 + 9, inf = 1e9 + 7;
ll infll = 1e18 + 7;

int n, k;
int a[N];

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    cin >> n >> k;
    ForE(i, 1, n){
        cin >> a[i];
    }
    ForE(i, k + 1, n){
        if (a[i] > a[i - k]) cout << "Yes";
        else cout << "No";
        cout << endl;
    }
}

/*
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*/