#include <bits/stdc++.h>
using namespace std;
int a[110000];
int b[110000];
int ta[110000];
int tb[110000];
int srt[110000];
int numa[210000];
int numb[210000];
int main() {
  int n, m, gg;
  scanf("%d %d %d", &n, &m, &gg);
  int num = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    srt[num++] = a[i];
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &b[i]);
    srt[num++] = b[i];
  }
  sort(srt, srt + num);
  num = unique(srt, srt + num) - srt;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(srt, srt + num, a[i]) - srt;
  }
  for (int i = 0; i < m; i++) {
    b[i] = lower_bound(srt, srt + num, b[i]) - srt;
  }
  sort(a, a + n);
  sort(b, b + m);
  int chk = 1;
  int nn = 0, mm = 0;
  for (int i = 0; i < n; i++) {
    int *ptr = lower_bound(b, b + m, a[i]);
    if (ptr == b + m) {
      printf("YES\n");
      return 0;
    }
    ta[nn++] = *ptr;
  }
  for (int i = 0; i < m; i++) {
    int ind = upper_bound(ta, ta + nn, b[i]) - 1 - ta;
    if (ind < 0) continue;
    tb[mm++] = ta[ind];
  }
  n = nn;
  m = mm;
  if (m == 0) {
    printf("YES\n");
    return 0;
  }
  sort(ta, ta + n);
  sort(tb, tb + m);
  gg = ta[n - 1];
  if (tb[m - 1] > gg) gg = tb[m - 1];
  gg++;
  for (int i = 0; i < n; i++) {
    numa[ta[i]]++;
  }
  for (int i = 0; i < m; i++) {
    numb[tb[i]]++;
  }
  for (int i = gg - 1; i >= 0; i--) {
    if (numa[i] > numb[i]) {
      printf("YES\n");
      return 0;
    } else {
      if (i - 1 >= 0) {
        numa[i - 1] += numa[i];
        numb[i - 1] += numb[i];
      }
    }
  }
  printf("NO\n");
}

### 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[110000];
int b[110000];
int ta[110000];
int tb[110000];
int srt[110000];
int numa[210000];
int numb[210000];
int main() {
  int n, m, gg;
  scanf("%d %d %d", &n, &m, &gg);
  int num = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    srt[num++] = a[i];
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &b[i]);
    srt[num++] = b[i];
  }
  sort(srt, srt + num);
  num = unique(srt, srt + num) - srt;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(srt, srt + num, a[i]) - srt;
  }
  for (int i = 0; i < m; i++) {
    b[i] = lower_bound(srt, srt + num, b[i]) - srt;
  }
  sort(a, a + n);
  sort(b, b + m);
  int chk = 1;
  int nn = 0, mm = 0;
  for (int i = 0; i < n; i++) {
    int *ptr = lower_bound(b, b + m, a[i]);
    if (ptr == b + m) {
      printf("YES\n");
      return 0;
    }
    ta[nn++] = *ptr;
  }
  for (int i = 0; i < m; i++) {
    int ind = upper_bound(ta, ta + nn, b[i]) - 1 - ta;
    if (ind < 0) continue;
    tb[mm++] = ta[ind];
  }
  n = nn;
  m = mm;
  if (m == 0) {
    printf("YES\n");
    return 0;
  }
  sort(ta, ta + n);
  sort(tb, tb + m);
  gg = ta[n - 1];
  if (tb[m - 1] > gg) gg = tb[m - 1];
  gg++;
  for (int i = 0; i < n; i++) {
    numa[ta[i]]++;
  }
  for (int i = 0; i < m; i++) {
    numb[tb[i]]++;
  }
  for (int i = gg - 1; i >= 0; i--) {
    if (numa[i] > numb[i]) {
      printf("YES\n");
      return 0;
    } else {
      if (i - 1 >= 0) {
        numa[i - 1] += numa[i];
        numb[i - 1] += numb[i];
      }
    }
  }
  printf("NO\n");
}

```