#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
bool isPrime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
bool isPalindrome(string s) {
  int l = 0;
  int h = s.size() - 1;
  while (h > l) {
    if (s[l++] != s[h--]) {
      return false;
    }
  }
  return true;
}
int ternarySearch(int l, int r, int key, int ar[]) {
  while (r >= l) {
    int mid1 = l + (r - l) / 3;
    int mid2 = r - (r - l) / 3;
    if (ar[mid1] == key) {
      return mid1;
    }
    if (ar[mid2] == key) {
      return mid2;
    }
    if (key < ar[mid1]) {
      r = mid1 - 1;
    } else if (key > ar[mid2]) {
      l = mid2 + 1;
    } else {
      l = mid1 + 1;
      r = mid2 - 1;
    }
  }
  return -1;
}
int binarySearch(int arr[], int l, int r, int x) {
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  int k, x;
  cin >> n >> m;
  cin >> k >> x;
  long long a[n + 1], b[m + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  if (a[k] < b[m - x + 1]) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long mod = 1000000007;
bool isPrime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
bool isPalindrome(string s) {
  int l = 0;
  int h = s.size() - 1;
  while (h > l) {
    if (s[l++] != s[h--]) {
      return false;
    }
  }
  return true;
}
int ternarySearch(int l, int r, int key, int ar[]) {
  while (r >= l) {
    int mid1 = l + (r - l) / 3;
    int mid2 = r - (r - l) / 3;
    if (ar[mid1] == key) {
      return mid1;
    }
    if (ar[mid2] == key) {
      return mid2;
    }
    if (key < ar[mid1]) {
      r = mid1 - 1;
    } else if (key > ar[mid2]) {
      l = mid2 + 1;
    } else {
      l = mid1 + 1;
      r = mid2 - 1;
    }
  }
  return -1;
}
int binarySearch(int arr[], int l, int r, int x) {
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  int k, x;
  cin >> n >> m;
  cin >> k >> x;
  long long a[n + 1], b[m + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  if (a[k] < b[m - x + 1]) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

```