#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void quickSort(long long int arr[], long long int left, long long int right) {
  long long int i = left, j = right;
  long long int tmp;
  long long int pivot = arr[(left + right) / 2];
  while (i <= j) {
    while (arr[i] < pivot) i++;
    while (arr[j] > pivot) j--;
    if (i <= j) {
      tmp = arr[i];
      arr[i] = arr[j];
      arr[j] = tmp;
      i++;
      j--;
    }
  };
  if (left < j) quickSort(arr, left, j);
  if (i < right) quickSort(arr, i, right);
}
int main() {
  long long int na, nb, i, k, m;
  cin >> na >> nb;
  cin >> k >> m;
  long long int a[na], b[nb];
  for (i = 0; i < na; i++) cin >> a[i];
  for (i = 0; i < nb; i++) cin >> b[i];
  if (b[nb - m] > a[k - 1])
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

### Prompt

Generate 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>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void quickSort(long long int arr[], long long int left, long long int right) {
  long long int i = left, j = right;
  long long int tmp;
  long long int pivot = arr[(left + right) / 2];
  while (i <= j) {
    while (arr[i] < pivot) i++;
    while (arr[j] > pivot) j--;
    if (i <= j) {
      tmp = arr[i];
      arr[i] = arr[j];
      arr[j] = tmp;
      i++;
      j--;
    }
  };
  if (left < j) quickSort(arr, left, j);
  if (i < right) quickSort(arr, i, right);
}
int main() {
  long long int na, nb, i, k, m;
  cin >> na >> nb;
  cin >> k >> m;
  long long int a[na], b[nb];
  for (i = 0; i < na; i++) cin >> a[i];
  for (i = 0; i < nb; i++) cin >> b[i];
  if (b[nb - m] > a[k - 1])
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```