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.
InputThe 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.
OutputPrint "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).
Examples 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 NoteIn 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:
题目大意：从两个数组中各取k个数，A数组中的数是否都小于B数组
思路：输入是按顺序的，就比较A的第k个和B的倒数k个就行
#include<iostream> #include<cstdio> #define manx 100005 using namespace std; int main() { int n,m,k,t; int a[manx]={0},b[manx]={0}; scanf("%d%d",&n,&m); scanf("%d%d",&k,&t); for (int i=1; i<=n; i++) scanf("%d",&a[i]); for (int i=1; i<=m; i++) scanf("%d",&b[i]); if(a[k] < b[m-t+1]) printf("YES\n"); else printf("NO\n"); return 0; }