#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int leap[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n, m, k;
map<int, int> M, M2;
vector<pair<int, int> > V;
int main() {
  while (scanf("%d%d%d", &n, &m, &k) == 3) {
    M.clear();
    M2.clear();
    V.clear();
    int p;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &p);
      M[p]++;
    }
    for (int i = 0; i < m; ++i) {
      scanf("%d", &p);
      M2[p]++;
    }
    bool bi = true;
    for (map<int, int>::iterator it = M.begin(); bi && it != M.end(); ++it)
      if (M2.find(it->first) == M2.end())
        V.push_back(make_pair(it->first, it->second));
      else if (it->second - M2[it->first] != 0)
        V.push_back(make_pair(it->first, it->second - M2[it->first]));
    for (map<int, int>::iterator it = M2.begin(); bi && it != M2.end(); ++it)
      if (M.find(it->first) == M.end())
        V.push_back(make_pair(it->first, -it->second));
    sort(V.begin(), V.end());
    int t = 0;
    for (int i = V.size() - 1; t <= 0 && i >= 0; --i) t += V[i].second;
    if (t > 0)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

### 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;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int leap[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n, m, k;
map<int, int> M, M2;
vector<pair<int, int> > V;
int main() {
  while (scanf("%d%d%d", &n, &m, &k) == 3) {
    M.clear();
    M2.clear();
    V.clear();
    int p;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &p);
      M[p]++;
    }
    for (int i = 0; i < m; ++i) {
      scanf("%d", &p);
      M2[p]++;
    }
    bool bi = true;
    for (map<int, int>::iterator it = M.begin(); bi && it != M.end(); ++it)
      if (M2.find(it->first) == M2.end())
        V.push_back(make_pair(it->first, it->second));
      else if (it->second - M2[it->first] != 0)
        V.push_back(make_pair(it->first, it->second - M2[it->first]));
    for (map<int, int>::iterator it = M2.begin(); bi && it != M2.end(); ++it)
      if (M.find(it->first) == M.end())
        V.push_back(make_pair(it->first, -it->second));
    sort(V.begin(), V.end());
    int t = 0;
    for (int i = V.size() - 1; t <= 0 && i >= 0; --i) t += V[i].second;
    if (t > 0)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

```