#include <bits/stdc++.h>
using namespace std;
int alice[200200], bob[200200];
vector<int> a, b;
int types[200200];
int main() {
  int N, M, K;
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 0; i < N; i++) {
    int in;
    scanf("%d", &in);
    a.push_back(in);
    types[i] = in;
  }
  for (int i = 0; i < M; i++) {
    int in;
    scanf("%d", &in);
    b.push_back(in);
    types[i + N] = in;
  }
  sort(types, types + N + M);
  int ln = unique(types, types + N + M) - types;
  for (int i = 0; i < N; i++) {
    int id = lower_bound(types, types + ln, a[i]) - types;
    alice[id]++;
  }
  for (int i = 0; i < M; i++) {
    int id = lower_bound(types, types + ln, b[i]) - types;
    bob[id]++;
  }
  int asum = 0, bsum = 0;
  for (int i = ln - 1; i >= 0; i--) {
    asum += alice[i], bsum += bob[i];
    if (asum > bsum) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  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;
int alice[200200], bob[200200];
vector<int> a, b;
int types[200200];
int main() {
  int N, M, K;
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 0; i < N; i++) {
    int in;
    scanf("%d", &in);
    a.push_back(in);
    types[i] = in;
  }
  for (int i = 0; i < M; i++) {
    int in;
    scanf("%d", &in);
    b.push_back(in);
    types[i + N] = in;
  }
  sort(types, types + N + M);
  int ln = unique(types, types + N + M) - types;
  for (int i = 0; i < N; i++) {
    int id = lower_bound(types, types + ln, a[i]) - types;
    alice[id]++;
  }
  for (int i = 0; i < M; i++) {
    int id = lower_bound(types, types + ln, b[i]) - types;
    bob[id]++;
  }
  int asum = 0, bsum = 0;
  for (int i = ln - 1; i >= 0; i--) {
    asum += alice[i], bsum += bob[i];
    if (asum > bsum) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}

```