#include <bits/stdc++.h>
using namespace std;
const int MAX = 100005;
int n, m, k;
int d1[MAX], d2[MAX];
int Table1[MAX * 2], Table2[MAX * 2];
vector<int> d;
bool check;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d1[i]);
    d.push_back(d1[i]);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &d2[i]);
    d.push_back(d2[i]);
  }
  sort(d1 + 1, d1 + 1 + n);
  sort(d2 + 1, d2 + 1 + m);
  sort(d.begin(), d.end());
  d.erase(unique(d.begin(), d.end()), d.end());
  int st = 0;
  for (int i = 1; i <= m; i++) {
    if (d2[i] >= d1[n])
      d2[i] = d1[n];
    else if (d2[i] < d1[1])
      d2[i] = 0;
    else {
      int inx;
      inx = lower_bound(d1 + 1, d1 + 1 + n, d2[i]) - d1;
      if (d2[i] < d1[inx])
        d2[i] = d1[inx - 1];
      else
        d2[i] = d1[inx];
    }
  }
  for (int i = 1; i <= n; i++)
    d1[i] = lower_bound(d.begin(), d.end(), d1[i]) - d.begin() + 1;
  for (int i = 1; i <= m; i++) {
    if (d2[i] != 0) {
      d2[i] = lower_bound(d.begin(), d.end(), d2[i]) - d.begin() + 1;
    }
  }
  for (int i = 1; i <= n; i++) Table1[d1[i]]++;
  for (int i = 1; i <= m; i++) Table2[d2[i]]++;
  for (int i = MAX * 2 - 1; i >= 1; i--) {
    if (Table1[i] > Table2[i]) {
      printf("YES\n");
      return 0;
    }
    Table1[i - 1] += Table1[i];
    Table2[i - 1] += Table2[i];
  }
  if (Table1[0] > Table2[0]) {
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}

### Prompt

Generate 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 int MAX = 100005;
int n, m, k;
int d1[MAX], d2[MAX];
int Table1[MAX * 2], Table2[MAX * 2];
vector<int> d;
bool check;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d1[i]);
    d.push_back(d1[i]);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &d2[i]);
    d.push_back(d2[i]);
  }
  sort(d1 + 1, d1 + 1 + n);
  sort(d2 + 1, d2 + 1 + m);
  sort(d.begin(), d.end());
  d.erase(unique(d.begin(), d.end()), d.end());
  int st = 0;
  for (int i = 1; i <= m; i++) {
    if (d2[i] >= d1[n])
      d2[i] = d1[n];
    else if (d2[i] < d1[1])
      d2[i] = 0;
    else {
      int inx;
      inx = lower_bound(d1 + 1, d1 + 1 + n, d2[i]) - d1;
      if (d2[i] < d1[inx])
        d2[i] = d1[inx - 1];
      else
        d2[i] = d1[inx];
    }
  }
  for (int i = 1; i <= n; i++)
    d1[i] = lower_bound(d.begin(), d.end(), d1[i]) - d.begin() + 1;
  for (int i = 1; i <= m; i++) {
    if (d2[i] != 0) {
      d2[i] = lower_bound(d.begin(), d.end(), d2[i]) - d.begin() + 1;
    }
  }
  for (int i = 1; i <= n; i++) Table1[d1[i]]++;
  for (int i = 1; i <= m; i++) Table2[d2[i]]++;
  for (int i = MAX * 2 - 1; i >= 1; i--) {
    if (Table1[i] > Table2[i]) {
      printf("YES\n");
      return 0;
    }
    Table1[i - 1] += Table1[i];
    Table2[i - 1] += Table2[i];
  }
  if (Table1[0] > Table2[0]) {
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}

```