#include <bits/stdc++.h>
using namespace std;
bool is_prime(long long int);
long long int gcd(long long int, long long int);
set<long long int> divs(long long int);
void swap(long long int& x, long long int& y) {
  long long int tmp = x;
  x = y;
  y = tmp;
}
const long long int N = 1000;
long long int cols[N], rows[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int n, x, y;
  cin >> n >> x >> y;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int r = 0;
  set<long long int> pfx, sfx;
  for (long long int i = 0; i < y; i++) sfx.insert(arr[min(i, n - 1)]);
  for (long long int i = 0; i < n; i++) {
    if (i - x > 0) {
      pfx.insert(arr[i - x]);
      pfx.erase(arr[i - x - 1]);
    } else {
      if (i) pfx.insert(arr[i - 1]);
    }
    if (i + y < n) {
      sfx.insert(arr[i + y]);
      sfx.erase(arr[i]);
    } else {
      if (sfx.size()) sfx.erase(arr[i]);
    }
    if (!pfx.size() || arr[i] <= *pfx.begin()) {
      if (!sfx.size() || arr[i] <= *sfx.begin()) {
        r = i + 1;
        break;
      }
    }
  }
  cout << r;
}
bool is_prime(long long int n) {
  if (n < 2) return 0;
  if (n == 2) return 1;
  for (long long int i = 2; i <= (long long int)sqrt(n); i++) {
    if ((n % i) == 0) return 0;
  }
  return 1;
}
long long int gcd(long long int x, long long int y) {
  return y == 0 ? x : gcd(y, x % y);
}
set<long long int> divs(long long int n) {
  set<long long int> st;
  for (long long int i = 1; i <= (long long int)sqrt(n); i++) {
    if (n % i == 0) {
      st.insert(i);
      st.insert(n / i);
    }
  }
  return st;
}

### Prompt

Generate 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;
bool is_prime(long long int);
long long int gcd(long long int, long long int);
set<long long int> divs(long long int);
void swap(long long int& x, long long int& y) {
  long long int tmp = x;
  x = y;
  y = tmp;
}
const long long int N = 1000;
long long int cols[N], rows[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int n, x, y;
  cin >> n >> x >> y;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int r = 0;
  set<long long int> pfx, sfx;
  for (long long int i = 0; i < y; i++) sfx.insert(arr[min(i, n - 1)]);
  for (long long int i = 0; i < n; i++) {
    if (i - x > 0) {
      pfx.insert(arr[i - x]);
      pfx.erase(arr[i - x - 1]);
    } else {
      if (i) pfx.insert(arr[i - 1]);
    }
    if (i + y < n) {
      sfx.insert(arr[i + y]);
      sfx.erase(arr[i]);
    } else {
      if (sfx.size()) sfx.erase(arr[i]);
    }
    if (!pfx.size() || arr[i] <= *pfx.begin()) {
      if (!sfx.size() || arr[i] <= *sfx.begin()) {
        r = i + 1;
        break;
      }
    }
  }
  cout << r;
}
bool is_prime(long long int n) {
  if (n < 2) return 0;
  if (n == 2) return 1;
  for (long long int i = 2; i <= (long long int)sqrt(n); i++) {
    if ((n % i) == 0) return 0;
  }
  return 1;
}
long long int gcd(long long int x, long long int y) {
  return y == 0 ? x : gcd(y, x % y);
}
set<long long int> divs(long long int n) {
  set<long long int> st;
  for (long long int i = 1; i <= (long long int)sqrt(n); i++) {
    if (n % i == 0) {
      st.insert(i);
      st.insert(n / i);
    }
  }
  return st;
}

```