[CF572A]Arrays

A. Arrays

 

You are given two arrays A and B consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose k numbers in array A and choose m numbers in array B so that any number chosen in the first array is strictly less than any number chosen in the second array.

Input

The first line contains two integers n_A, n_B (1 ≤ n_A, n_B ≤ 10⁵), separated by a space — the sizes of arrays A and B, correspondingly.

The second line contains two integers k and m (1 ≤ k ≤ n_A, 1 ≤ m ≤ n_B), separated by a space.

The third line contains n_A numbers a₁, a₂, ... a_nA ( - 10⁹ ≤ a₁ ≤ a₂ ≤ ... ≤ a_nA ≤ 10⁹), separated by spaces — elements of array A.

The fourth line contains n_B integers b₁, b₂, ... b_nB ( - 10⁹ ≤ b₁ ≤ b₂ ≤ ... ≤ b_nB ≤ 10⁹), separated by spaces — elements of array B.

Output

Print "YES" (without the quotes), if you can choose k numbers in array A and m numbers in array B so that any number chosen in array A was strictly less than any number chosen in array B. Otherwise, print "NO" (without the quotes).

Sample test(s)

Input

3 3
2 1
1 2 3
3 4 5

Output

YES

Input

3 3
3 3
1 2 3
3 4 5

Output

NO

Input

5 2
3 1
1 1 1 1 1
2 2

Output

YES

Note

In the first sample test you can, for example, choose numbers 1 and 2 from array A and number 3 from array B (1 < 3 and 2 < 3).

In the second sample test the only way to choose k elements in the first array and m elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in A will be less than all the numbers chosen in B: .

直接比较a的第k个和b的第nb-m+1个元素的大小。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#include <ctime>

using namespace std;

const int maxn = 100010;

int a[maxn];
int b[maxn];
int k, m;
int na, nb;

int main() {
    while(~scanf("%d%d", &na, &nb)) {
        scanf("%d%d", &k, &m);
        for(int i = 1; i <= na; i++) {
            scanf("%d", &a[i]);
        }
        for(int i = 1; i <= nb; i++) {
            scanf("%d", &b[i]);
        }
        if(a[k] < b[nb-m+1]) {
            printf("YES
");
        }
        else {
            printf("NO
");
        }
    }
}