#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> a(n), b(m);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  map<int, int> z;
  int j = 0;
  for (int i = 0; i < n; ++i) {
    ++z[a[i]];
    while (j < m && (i == n - 1 || b[j] < a[i + 1])) {
      if (b[j] >= a[i]) {
        --z[a[i]];
      }
      ++j;
    }
  }
  vector<int> res;
  for (map<int, int>::iterator it = z.begin(); it != z.end(); ++it) {
    res.push_back(it->second);
  }
  int r = 0;
  for (int i = res.size() - 1; i >= 0; --i) {
    r += res[i];
    if (r > 0) {
      cout << "YES\n";
      return 0;
    }
  }
  cout << "NO\n";
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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;
  cin >> n >> m >> k;
  vector<int> a(n), b(m);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  map<int, int> z;
  int j = 0;
  for (int i = 0; i < n; ++i) {
    ++z[a[i]];
    while (j < m && (i == n - 1 || b[j] < a[i + 1])) {
      if (b[j] >= a[i]) {
        --z[a[i]];
      }
      ++j;
    }
  }
  vector<int> res;
  for (map<int, int>::iterator it = z.begin(); it != z.end(); ++it) {
    res.push_back(it->second);
  }
  int r = 0;
  for (int i = res.size() - 1; i >= 0; --i) {
    r += res[i];
    if (r > 0) {
      cout << "YES\n";
      return 0;
    }
  }
  cout << "NO\n";
  return 0;
}

```