#include <bits/stdc++.h>
using namespace std;
const long long md = 1e9 + 7;
const long long MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power_md(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res % md * a % md) % md;
    a = (a % md * a % md) % md;
    n /= 2;
  }
  res %= md;
  return res;
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a = a * a;
    n /= 2;
  }
  return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
class cmp_set {
 public:
  bool operator()(long long a, long long b) { return a > b; }
};
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> a;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    a.push_back(x);
  }
  vector<long long> b;
  for (long long i = 1; i <= m; i++) {
    long long x;
    cin >> x;
    b.push_back(x);
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  if (n > m)
    cout << "YES\n";
  else {
    long long f = 0;
    for (long long i = a.size() - 1; i >= 0; i--) {
      long long c1 = b.end() - lower_bound(b.begin(), b.end(), a[i]);
      long long c = a.size() - i;
      if (c > c1) {
        f = 1;
        break;
      }
    }
    if (f)
      cout << "YES\n";
    else
      cout << "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 long long md = 1e9 + 7;
const long long MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power_md(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res % md * a % md) % md;
    a = (a % md * a % md) % md;
    n /= 2;
  }
  res %= md;
  return res;
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a = a * a;
    n /= 2;
  }
  return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
class cmp_set {
 public:
  bool operator()(long long a, long long b) { return a > b; }
};
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> a;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    a.push_back(x);
  }
  vector<long long> b;
  for (long long i = 1; i <= m; i++) {
    long long x;
    cin >> x;
    b.push_back(x);
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  if (n > m)
    cout << "YES\n";
  else {
    long long f = 0;
    for (long long i = a.size() - 1; i >= 0; i--) {
      long long c1 = b.end() - lower_bound(b.begin(), b.end(), a[i]);
      long long c = a.size() - i;
      if (c > c1) {
        f = 1;
        break;
      }
    }
    if (f)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

```