#include <bits/stdc++.h>
using namespace std;
template <typename tp>
inline void read(tp& x) {
  x = 0;
  register int f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  x *= f;
}
template <typename Tp>
inline void print(Tp x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 200010;
int n, x, y;
int a[maxn];
inline bool check1(int day) {
  int nn;
  if (day - x >= 1)
    nn = day - x;
  else
    nn = 1;
  if (day == 1) return 1;
  for (register int i = nn; i < day; ++i) {
    if (a[i] < a[i + 1]) return 0;
  }
  return 1;
}
inline bool check2(int day) {
  int nn;
  if (day + y <= n)
    nn = day + y;
  else
    nn = n;
  if (day == nn) return 1;
  for (register int i = day + 1; i <= nn; ++i) {
    if (a[i - 1] > a[i]) return 0;
  }
  return 1;
}
int main() {
  read(n), read(x), read(y);
  for (register int i = 1; i <= n; ++i) read(a[i]);
  for (register int i = 1; i <= n; i++) {
    bool flag = 0;
    int x1 = max(1, i - x);
    int x2 = min(n, i + y);
    for (register int j = x1; j <= x2; j++) {
      if (i == j) continue;
      if (a[j] > a[i])
        continue;
      else {
        flag = 1;
        break;
      }
    }
    if (flag == 0) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 tp>
inline void read(tp& x) {
  x = 0;
  register int f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  x *= f;
}
template <typename Tp>
inline void print(Tp x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 200010;
int n, x, y;
int a[maxn];
inline bool check1(int day) {
  int nn;
  if (day - x >= 1)
    nn = day - x;
  else
    nn = 1;
  if (day == 1) return 1;
  for (register int i = nn; i < day; ++i) {
    if (a[i] < a[i + 1]) return 0;
  }
  return 1;
}
inline bool check2(int day) {
  int nn;
  if (day + y <= n)
    nn = day + y;
  else
    nn = n;
  if (day == nn) return 1;
  for (register int i = day + 1; i <= nn; ++i) {
    if (a[i - 1] > a[i]) return 0;
  }
  return 1;
}
int main() {
  read(n), read(x), read(y);
  for (register int i = 1; i <= n; ++i) read(a[i]);
  for (register int i = 1; i <= n; i++) {
    bool flag = 0;
    int x1 = max(1, i - x);
    int x2 = min(n, i + y);
    for (register int j = x1; j <= x2; j++) {
      if (i == j) continue;
      if (a[j] > a[i])
        continue;
      else {
        flag = 1;
        break;
      }
    }
    if (flag == 0) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}

```