#include <bits/stdc++.h>
using namespace std;
int A[100666], B[100666], Nr[100666 * 2];
int fr1[100666 * 2], fr2[100666 * 2];
int bs(int value, int lmax) {
  int pos, step;
  for (step = 1; step <= lmax; step <<= 1)
    ;
  for (pos = 0; step; step >>= 1)
    if (pos + step <= lmax && Nr[pos + step] <= value) pos += step;
  return pos + step;
}
int main() {
  int n, m, k, i, tps = 1, nrel = 0;
  cin >> n >> m >> k;
  for (i = 1; i <= n; ++i) {
    cin >> A[i];
    Nr[++nrel] = A[i];
  }
  for (i = 1; i <= m; ++i) {
    cin >> B[i];
    Nr[++nrel] = B[i];
  }
  sort(Nr + 1, Nr + nrel + 1);
  for (i = 2; i <= nrel; ++i)
    if (Nr[i] != Nr[tps]) Nr[++tps] = Nr[i];
  for (i = 1; i <= n; ++i) ++fr1[bs(A[i], tps)];
  for (i = 1; i <= m; ++i) ++fr2[bs(B[i], tps)];
  int s1 = 0, s2 = 0;
  for (i = tps; i > 0; --i) {
    s1 += fr1[i];
    s2 += fr2[i];
    if (s1 > s2) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  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;
int A[100666], B[100666], Nr[100666 * 2];
int fr1[100666 * 2], fr2[100666 * 2];
int bs(int value, int lmax) {
  int pos, step;
  for (step = 1; step <= lmax; step <<= 1)
    ;
  for (pos = 0; step; step >>= 1)
    if (pos + step <= lmax && Nr[pos + step] <= value) pos += step;
  return pos + step;
}
int main() {
  int n, m, k, i, tps = 1, nrel = 0;
  cin >> n >> m >> k;
  for (i = 1; i <= n; ++i) {
    cin >> A[i];
    Nr[++nrel] = A[i];
  }
  for (i = 1; i <= m; ++i) {
    cin >> B[i];
    Nr[++nrel] = B[i];
  }
  sort(Nr + 1, Nr + nrel + 1);
  for (i = 2; i <= nrel; ++i)
    if (Nr[i] != Nr[tps]) Nr[++tps] = Nr[i];
  for (i = 1; i <= n; ++i) ++fr1[bs(A[i], tps)];
  for (i = 1; i <= m; ++i) ++fr2[bs(B[i], tps)];
  int s1 = 0, s2 = 0;
  for (i = tps; i > 0; --i) {
    s1 += fr1[i];
    s2 += fr2[i];
    if (s1 > s2) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}

```