#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, k, l;
  cin >> i >> j >> k;
  int A[i], B[j];
  for (l = 0; l < i; l++) cin >> A[l];
  for (l = 0; l < j; l++) cin >> B[l];
  sort(A, A + i);
  sort(B, B + j);
  if (i > j) {
    cout << "YES\n";
    return 0;
  }
  for (k = i - 1, l = j - 1; k >= 0 && l >= 0; k--, l--) {
    if (A[k] > B[l]) break;
  }
  if (k >= 0 && l >= 0)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

It is known that there are k fish species in the polar ocean, numbered from 1 to k. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the i-th type of fish be wi, then 0 < w1 ≤ w2 ≤ ... ≤ wk holds.

Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights wi (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?

Input

The first line contains three integers n, m, k (1 ≤ n, m ≤ 105, 1 ≤ k ≤ 109) — the number of fish caught by Alice and Bob respectively, and the number of fish species.

The second line contains n integers each from 1 to k, the list of fish type caught by Alice. The third line contains m integers each from 1 to k, the list of fish type caught by Bob.

Note that one may have caught more than one fish for a same species.

Output

Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.

Examples

Input

3 3 3
2 2 2
1 1 3


Output

YES


Input

4 7 9
5 2 7 3
3 5 2 7 3 8 7


Output

NO

Note

In the first sample, if w1 = 1, w2 = 2, w3 = 2.5, then Alice has a total of 2 + 2 + 2 = 6 weight units, while Bob only has 1 + 1 + 2.5 = 4.5.

In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, k, l;
  cin >> i >> j >> k;
  int A[i], B[j];
  for (l = 0; l < i; l++) cin >> A[l];
  for (l = 0; l < j; l++) cin >> B[l];
  sort(A, A + i);
  sort(B, B + j);
  if (i > j) {
    cout << "YES\n";
    return 0;
  }
  for (k = i - 1, l = j - 1; k >= 0 && l >= 0; k--, l--) {
    if (A[k] > B[l]) break;
  }
  if (k >= 0 && l >= 0)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

```