#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &X) {
  X = 0;
  char ch = 0;
  T op = 1;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') op = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    X = (X << 3) + (X << 1) + ch - 48;
  X *= op;
}
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
int a[N];
int main() {
  int n, x, y;
  read(n);
  cin >> x >> y;
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) {
    bool ok = true;
    for (int j = max(i - x, 1); j < i; j++) {
      if (a[j] <= a[i]) {
        ok = false;
        break;
      }
    }
    for (int j = i + 1; j <= min(i + y, n); j++) {
      if (a[j] <= a[i]) {
        ok = false;
        break;
      }
    }
    if (ok) {
      cout << i;
      return 0;
    }
  }
  return 0;
}

### Prompt

Please formulate 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 <typename T>
inline void read(T &X) {
  X = 0;
  char ch = 0;
  T op = 1;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') op = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    X = (X << 3) + (X << 1) + ch - 48;
  X *= op;
}
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
int a[N];
int main() {
  int n, x, y;
  read(n);
  cin >> x >> y;
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) {
    bool ok = true;
    for (int j = max(i - x, 1); j < i; j++) {
      if (a[j] <= a[i]) {
        ok = false;
        break;
      }
    }
    for (int j = i + 1; j <= min(i + y, n); j++) {
      if (a[j] <= a[i]) {
        ok = false;
        break;
      }
    }
    if (ok) {
      cout << i;
      return 0;
    }
  }
  return 0;
}

```