#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return SQR(POW(B, P / 2));
}
template <typename T>
inline T ModInv(T b, T m) {
  return BigMod(b, m - 2, m);
}
template <typename T>
inline T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <class T>
inline T BMOD(T p, T e, T m) {
  T ret = 1;
  while (e) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
    e >>= 1;
  }
  return (T)ret;
}
int a[100005], b[100005], na, nb, k, m;
int main() {
  cin >> na >> nb;
  cin >> k >> m;
  for (int i = 1; i <= na; i++) cin >> a[i];
  for (int i = 1; i <= nb; i++) cin >> b[i];
  int x = b[nb - m + 1];
  int cnt = 0;
  for (int i = 1; i <= na; i++) {
    if (a[i] < x)
      cnt++;
    else
      break;
    if (cnt == k) break;
  }
  if (cnt >= k)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
template <typename T>
inline T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return SQR(POW(B, P / 2));
}
template <typename T>
inline T ModInv(T b, T m) {
  return BigMod(b, m - 2, m);
}
template <typename T>
inline T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <class T>
inline T BMOD(T p, T e, T m) {
  T ret = 1;
  while (e) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
    e >>= 1;
  }
  return (T)ret;
}
int a[100005], b[100005], na, nb, k, m;
int main() {
  cin >> na >> nb;
  cin >> k >> m;
  for (int i = 1; i <= na; i++) cin >> a[i];
  for (int i = 1; i <= nb; i++) cin >> b[i];
  int x = b[nb - m + 1];
  int cnt = 0;
  for (int i = 1; i <= na; i++) {
    if (a[i] < x)
      cnt++;
    else
      break;
    if (cnt == k) break;
  }
  if (cnt >= k)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```