#include <bits/stdc++.h>
const int inf = (1ll << 31) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
int n1 = 0, n2 = 0, dat1[110000], dat2[110000];
int k = 0, m = 0;
int main() {
  while (scanf("%d%d", &n1, &n2) != EOF) {
    scanf("%d%d", &k, &m);
    for (int i = 0; i < n1; i++) {
      cin >> dat1[i];
    }
    for (int i = 0; i < n2; i++) {
      cin >> dat2[i];
    }
    int add1 = 0, add2 = 0;
    int first = 1;
    int minn2 = inf, maxn2 = -inf;
    for (int j = n2 - 1; j >= n2 - m; j--) {
      if (minn2 > dat2[j]) {
        minn2 = dat2[j];
      }
      if (maxn2 < dat2[j]) {
        maxn2 = dat2[j];
      }
    }
    for (int i = 0; i < k; i++) {
      if (dat1[i] >= minn2) {
        first = 0;
      }
    }
    if (first) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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 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 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

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: <image>.

### Response

```cpp
#include <bits/stdc++.h>
const int inf = (1ll << 31) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
int n1 = 0, n2 = 0, dat1[110000], dat2[110000];
int k = 0, m = 0;
int main() {
  while (scanf("%d%d", &n1, &n2) != EOF) {
    scanf("%d%d", &k, &m);
    for (int i = 0; i < n1; i++) {
      cin >> dat1[i];
    }
    for (int i = 0; i < n2; i++) {
      cin >> dat2[i];
    }
    int add1 = 0, add2 = 0;
    int first = 1;
    int minn2 = inf, maxn2 = -inf;
    for (int j = n2 - 1; j >= n2 - m; j--) {
      if (minn2 > dat2[j]) {
        minn2 = dat2[j];
      }
      if (maxn2 < dat2[j]) {
        maxn2 = dat2[j];
      }
    }
    for (int i = 0; i < k; i++) {
      if (dat1[i] >= minn2) {
        first = 0;
      }
    }
    if (first) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}

```