#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / (gcd(a, b)));
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T1>
void deb(T1 e) {
  cout << e << endl;
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
  cout << e1 << " " << e2 << endl;
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
  cout << e1 << " " << e2 << " " << e3 << endl;
}
template <class T1, class T2, class T3, class T4>
void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << endl;
}
template <class T1, class T2, class T3, class T4, class T5>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << endl;
}
template <class T1, class T2, class T3, class T4, class T5, class T6>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
       << endl;
}
int main() {
  int i, j, x, y, a, d, n, res = -1;
  int days[100010];
  n = ({
    int a;
    scanf("%d", &a);
    a;
  });
  x = ({
    int a;
    scanf("%d", &a);
    a;
  });
  y = ({
    int a;
    scanf("%d", &a);
    a;
  });
  for (i = 1; i <= n; i++) {
    a = ({
      int a;
      scanf("%d", &a);
      a;
    });
    days[i] = a;
  }
  for (i = 1; i <= n; i++) {
    bool success = true;
    for (j = i; j <= max(1, i - x); j++) {
      if (days[j] < days[i]) success = false;
    }
    if (!success) continue;
    for (j = i; j <= min(i + y, n); j++) {
      if (days[j] < days[i]) success = false;
    }
    if (!success)
      continue;
    else {
      res = i;
      break;
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / (gcd(a, b)));
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T1>
void deb(T1 e) {
  cout << e << endl;
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
  cout << e1 << " " << e2 << endl;
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
  cout << e1 << " " << e2 << " " << e3 << endl;
}
template <class T1, class T2, class T3, class T4>
void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << endl;
}
template <class T1, class T2, class T3, class T4, class T5>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << endl;
}
template <class T1, class T2, class T3, class T4, class T5, class T6>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
       << endl;
}
int main() {
  int i, j, x, y, a, d, n, res = -1;
  int days[100010];
  n = ({
    int a;
    scanf("%d", &a);
    a;
  });
  x = ({
    int a;
    scanf("%d", &a);
    a;
  });
  y = ({
    int a;
    scanf("%d", &a);
    a;
  });
  for (i = 1; i <= n; i++) {
    a = ({
      int a;
      scanf("%d", &a);
      a;
    });
    days[i] = a;
  }
  for (i = 1; i <= n; i++) {
    bool success = true;
    for (j = i; j <= max(1, i - x); j++) {
      if (days[j] < days[i]) success = false;
    }
    if (!success) continue;
    for (j = i; j <= min(i + y, n); j++) {
      if (days[j] < days[i]) success = false;
    }
    if (!success)
      continue;
    else {
      res = i;
      break;
    }
  }
  cout << res << endl;
  return 0;
}

```