#include <bits/stdc++.h>
using namespace std;
struct C {
  int num;
  int mark;
} c[200010];
map<int, int> Tb;
int cal[200010], n, m, k;
bool cmp(C a, C b) { return a.num < b.num; }
int main() {
  Tb.clear();
  memset(cal, 0, sizeof(cal));
  cin >> n >> m >> k;
  int cnt = 0, i;
  for (i = 0; i < n; i++) {
    cin >> c[i].num;
    c[i].mark = 0;
  }
  for (i = 0; i < m; i++) {
    cin >> c[i + n].num;
    c[i + n].mark = 1;
  }
  sort(c, c + m + n, cmp);
  for (i = 0; i < m + n; i++) {
    if (!Tb.count(c[i].num)) Tb[c[i].num] = cnt++;
    if (c[i].mark == 0)
      cal[Tb[c[i].num]]++;
    else
      cal[Tb[c[i].num]]--;
  }
  int sum = 0;
  bool mark = false;
  for (i = 0; i < cnt; i++) {
    if (cal[i] > 0 && sum < 0)
      sum = cal[i] - 1, mark = true;
    else
      sum += cal[i];
  }
  if (sum > 0 || (sum == 0 && mark))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### Prompt

Your task is to create 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;
struct C {
  int num;
  int mark;
} c[200010];
map<int, int> Tb;
int cal[200010], n, m, k;
bool cmp(C a, C b) { return a.num < b.num; }
int main() {
  Tb.clear();
  memset(cal, 0, sizeof(cal));
  cin >> n >> m >> k;
  int cnt = 0, i;
  for (i = 0; i < n; i++) {
    cin >> c[i].num;
    c[i].mark = 0;
  }
  for (i = 0; i < m; i++) {
    cin >> c[i + n].num;
    c[i + n].mark = 1;
  }
  sort(c, c + m + n, cmp);
  for (i = 0; i < m + n; i++) {
    if (!Tb.count(c[i].num)) Tb[c[i].num] = cnt++;
    if (c[i].mark == 0)
      cal[Tb[c[i].num]]++;
    else
      cal[Tb[c[i].num]]--;
  }
  int sum = 0;
  bool mark = false;
  for (i = 0; i < cnt; i++) {
    if (cal[i] > 0 && sum < 0)
      sum = cal[i] - 1, mark = true;
    else
      sum += cal[i];
  }
  if (sum > 0 || (sum == 0 && mark))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```