#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long int N = 1e5 + 5;
const long long int INF = 1e13;
long long int mod = 998244353;
long long int dx[4] = {0, 0, +1, -1};
long long int dy[4] = {+1, -1, 0, 0};
void solve() {
  long long int n, x, y;
  cin >> n >> x >> y;
  vector<long long int> v(n + 1, -1);
  for (long long int i = 1; i <= n; i++) cin >> v[i];
  for (long long int i = 1; i <= n; i++) {
    long long int f = 0;
    for (long long int j = i - 1; j >= max(i - x, (long long int)1); j--) {
      if (v[j] < v[i] and v[j] != -1) {
        f = 1;
        break;
      }
    }
    for (long long int j = i + 1; j <= min(n, i + y); j++) {
      if (v[j] < v[i] and v[j] != -1) {
        f = 1;
        break;
      }
    }
    if (f == 0) {
      cout << i << "\n";
      return;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.

The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.

Help mayor find the earliest not-so-rainy day of summer.

Input

The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.

The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.

Output

Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.

Examples

Input


10 2 2
10 9 6 7 8 3 2 1 4 5


Output


3


Input


10 2 3
10 9 6 7 8 3 2 1 4 5


Output


8


Input


5 5 5
100000 10000 1000 100 10


Output


5

Note

In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.

In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long int N = 1e5 + 5;
const long long int INF = 1e13;
long long int mod = 998244353;
long long int dx[4] = {0, 0, +1, -1};
long long int dy[4] = {+1, -1, 0, 0};
void solve() {
  long long int n, x, y;
  cin >> n >> x >> y;
  vector<long long int> v(n + 1, -1);
  for (long long int i = 1; i <= n; i++) cin >> v[i];
  for (long long int i = 1; i <= n; i++) {
    long long int f = 0;
    for (long long int j = i - 1; j >= max(i - x, (long long int)1); j--) {
      if (v[j] < v[i] and v[j] != -1) {
        f = 1;
        break;
      }
    }
    for (long long int j = i + 1; j <= min(n, i + y); j++) {
      if (v[j] < v[i] and v[j] != -1) {
        f = 1;
        break;
      }
    }
    if (f == 0) {
      cout << i << "\n";
      return;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  while (t--) solve();
  return 0;
}

```