#include <bits/stdc++.h>
using namespace std;
int main() {
  int N, M, K;
  scanf("%d %d %d", &N, &M, &K);
  map<int, int> a, b;
  for (int i = 0, t = 0; i < N; i++) scanf("%d", &t), a[t] += 1;
  for (int i = 0, t = 0; i < M; i++) scanf("%d", &t), b[t] += 1;
  map<int, int>::iterator ita = a.end(), itb = b.end();
  ita--;
  itb--;
  if (ita->first > itb->first ||
      (ita->first == itb->first && ita->second > itb->second)) {
    cout << "YES" << endl;
    return 0;
  }
  int t1 = 0, t2 = 0;
  bool b1 = true;
  while (t1 <= t2) {
    t1 += ita->second;
    while (itb->first >= ita->first && b1) {
      t2 += itb->second;
      if (itb == b.begin()) {
        b1 = false;
        break;
      }
      itb--;
    }
    if (ita == a.begin()) break;
    ita--;
  }
  if (t1 > t2)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### 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 N, M, K;
  scanf("%d %d %d", &N, &M, &K);
  map<int, int> a, b;
  for (int i = 0, t = 0; i < N; i++) scanf("%d", &t), a[t] += 1;
  for (int i = 0, t = 0; i < M; i++) scanf("%d", &t), b[t] += 1;
  map<int, int>::iterator ita = a.end(), itb = b.end();
  ita--;
  itb--;
  if (ita->first > itb->first ||
      (ita->first == itb->first && ita->second > itb->second)) {
    cout << "YES" << endl;
    return 0;
  }
  int t1 = 0, t2 = 0;
  bool b1 = true;
  while (t1 <= t2) {
    t1 += ita->second;
    while (itb->first >= ita->first && b1) {
      t2 += itb->second;
      if (itb == b.begin()) {
        b1 = false;
        break;
      }
      itb--;
    }
    if (ita == a.begin()) break;
    ita--;
  }
  if (t1 > t2)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```