Arrays

A. Arrays

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two arrays A and B consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose knumbers 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 nA, nB (1 ≤ nA, nB ≤ 105), separated by a space — the sizes of arrays A and B, correspondingly.

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

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

The fourth line contains nB integers b1, b2, ... bnB ( - 109 ≤ b1 ≤ b2 ≤ ... ≤ bnB ≤ 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 arrayA was strictly less than any number chosen in array B. Otherwise, print "NO" (without the quotes).

Sample test(s)

input

3 32 11 2 33 4 5

output

YES

input

3 33 31 2 33 4 5

output

NO

input

5 23 11 1 1 1 12 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: .

思路：

    一开始还以为是数组a的0-Na中最大的必须要小于数组b里的所有数。后来wa了，就又重新看题目。终于知道他是找b中的最小一段中的最大的那个数。

ac代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>using namespace std;typedef __int64 ll;#define INF 2000000000#define T 100100int ar1[T],ar2[T];int main(){   /* freopen("input.txt","r",stdin);*/    int n,i,m,a,b;    while(~scanf("%d%d",&n,&m))    {        scanf("%d%d",&a,&b);        for(i=0;i<n;++i){            scanf("%d",&ar1[i]);        }        for(i=0;i<m;++i){            scanf("%d",&ar2[i]);        }        sort(ar1,ar1+n);        sort(ar2,ar2+m);        if(ar1[a-1]>=ar2[m-b])            printf("NO\n");        else            printf("YES\n");    }    return 0;}