#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, y;
  cin >> n >> x >> y;
  vector<long long> v(n + 1);
  for (long long i = 1; i <= (n); ++i) cin >> v[i];
  int counter = 1;
  if (x == 0) {
    if (y == 0) {
      cout << 1 << '\n';
      return 0;
    } else {
      for (long long i = 1; i <= (n); ++i) {
        int i2 = i + 1;
        while (i2 <= min(i + y, n)) {
          counter = 1;
          if (v[i2] <= v[i]) {
            counter = 0;
            break;
          }
          i2++;
        }
        if (counter == 1) {
          cout << i << '\n';
          return 0;
        }
      }
    }
  }
  for (long long i = 1; i <= (n); ++i) {
    counter = 1;
    int i1 = i - 1, i2 = i + 1;
    while (i1 >= max(i - x, 1ll)) {
      if (v[i1] <= v[i]) {
        counter = 0;
        break;
      }
      i1--;
    }
    while (i2 <= min((i + y), n)) {
      if (v[i2] <= v[i]) {
        counter = 0;
        break;
      }
      i2++;
    }
    if (counter == 1) {
      cout << i << '\n';
      return 0;
    }
  }
  cout << n << '\n';
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, y;
  cin >> n >> x >> y;
  vector<long long> v(n + 1);
  for (long long i = 1; i <= (n); ++i) cin >> v[i];
  int counter = 1;
  if (x == 0) {
    if (y == 0) {
      cout << 1 << '\n';
      return 0;
    } else {
      for (long long i = 1; i <= (n); ++i) {
        int i2 = i + 1;
        while (i2 <= min(i + y, n)) {
          counter = 1;
          if (v[i2] <= v[i]) {
            counter = 0;
            break;
          }
          i2++;
        }
        if (counter == 1) {
          cout << i << '\n';
          return 0;
        }
      }
    }
  }
  for (long long i = 1; i <= (n); ++i) {
    counter = 1;
    int i1 = i - 1, i2 = i + 1;
    while (i1 >= max(i - x, 1ll)) {
      if (v[i1] <= v[i]) {
        counter = 0;
        break;
      }
      i1--;
    }
    while (i2 <= min((i + y), n)) {
      if (v[i2] <= v[i]) {
        counter = 0;
        break;
      }
      i2++;
    }
    if (counter == 1) {
      cout << i << '\n';
      return 0;
    }
  }
  cout << n << '\n';
  return 0;
}

```