#include <bits/stdc++.h>
const int N = 1e5 + 10;
using namespace std;
const long long int MOOD = 998244353;
bool sortBySec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
bool sortcol(const vector<long long int> &v1, const vector<long long int> &v2) {
  return (v1[0] < v2[0]);
}
long long int power(long long int a, long long int b) {
  long long int c = 1;
  while (b) {
    if (b & 1) c = ((long long int)c * a) % 1000000007;
    a = ((long long int)a * a) % 1000000007;
    b = b / 2;
    cout << "C    =   " << c << "    A    =   " << a << "    B    =   " << b
         << endl;
  }
  return c;
}
vector<pair<int, int> > graph[N];
int freq[N];
bool used[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    long long int n, x, y;
    cin >> n >> x >> y;
    long long int a[n];
    for (auto &x : a) {
      cin >> x;
    };
    long long int pos;
    if (y == 0) {
      cout << "1" << endl;
      return 0;
    }
    if (x == 0) {
      long long int pos = -1;
      for (long long int i = 0; i < n - 1; i++) {
        if (a[i] < a[i + 1]) {
          long long int temp = 0;
          for (long long int j = 0; j < y; j++) {
            long long int d = i + 1 + j;
            if (d > n - 1)
              break;
            else if (a[i] >= a[d]) {
              temp = 1;
              break;
            }
          }
          if (temp == 0) {
            pos = i + 1;
            break;
          }
        }
      }
      if (pos == -1)
        cout << n << endl;
      else
        cout << pos << endl;
      return 0;
    }
    for (long long int i = 0; i < n; i++) {
      if (i == 0) {
        if (a[i] < a[i + 1]) {
          long long int temp = 0;
          for (long long int j = 0; j < y; j++) {
            if (a[i + 1 + j] <= a[i]) {
              temp = 1;
              break;
            }
          }
          if (temp == 0) {
            pos = i + 1;
            break;
          }
        }
      } else if (i == n - 1) {
        pos = n;
      } else {
        if (a[i] < a[i - 1] && a[i] < a[i + 1]) {
          long long int temp = 0;
          for (long long int j = 0; j < x; j++) {
            long long int d = i - 1 - j;
            if (d < 0)
              break;
            else if (a[i] >= a[d]) {
              temp = 1;
              break;
            }
          }
          for (long long int j = 0; j < y; j++) {
            long long int d = i + 1 + j;
            if (d > n - 1) break;
            if (a[d] <= a[i]) {
              temp = 1;
              break;
            }
          }
          if (temp == 0) {
            pos = i + 1;
            break;
          }
        }
      }
    }
    cout << pos << endl;
  }
}

### 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>
const int N = 1e5 + 10;
using namespace std;
const long long int MOOD = 998244353;
bool sortBySec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
bool sortcol(const vector<long long int> &v1, const vector<long long int> &v2) {
  return (v1[0] < v2[0]);
}
long long int power(long long int a, long long int b) {
  long long int c = 1;
  while (b) {
    if (b & 1) c = ((long long int)c * a) % 1000000007;
    a = ((long long int)a * a) % 1000000007;
    b = b / 2;
    cout << "C    =   " << c << "    A    =   " << a << "    B    =   " << b
         << endl;
  }
  return c;
}
vector<pair<int, int> > graph[N];
int freq[N];
bool used[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    long long int n, x, y;
    cin >> n >> x >> y;
    long long int a[n];
    for (auto &x : a) {
      cin >> x;
    };
    long long int pos;
    if (y == 0) {
      cout << "1" << endl;
      return 0;
    }
    if (x == 0) {
      long long int pos = -1;
      for (long long int i = 0; i < n - 1; i++) {
        if (a[i] < a[i + 1]) {
          long long int temp = 0;
          for (long long int j = 0; j < y; j++) {
            long long int d = i + 1 + j;
            if (d > n - 1)
              break;
            else if (a[i] >= a[d]) {
              temp = 1;
              break;
            }
          }
          if (temp == 0) {
            pos = i + 1;
            break;
          }
        }
      }
      if (pos == -1)
        cout << n << endl;
      else
        cout << pos << endl;
      return 0;
    }
    for (long long int i = 0; i < n; i++) {
      if (i == 0) {
        if (a[i] < a[i + 1]) {
          long long int temp = 0;
          for (long long int j = 0; j < y; j++) {
            if (a[i + 1 + j] <= a[i]) {
              temp = 1;
              break;
            }
          }
          if (temp == 0) {
            pos = i + 1;
            break;
          }
        }
      } else if (i == n - 1) {
        pos = n;
      } else {
        if (a[i] < a[i - 1] && a[i] < a[i + 1]) {
          long long int temp = 0;
          for (long long int j = 0; j < x; j++) {
            long long int d = i - 1 - j;
            if (d < 0)
              break;
            else if (a[i] >= a[d]) {
              temp = 1;
              break;
            }
          }
          for (long long int j = 0; j < y; j++) {
            long long int d = i + 1 + j;
            if (d > n - 1) break;
            if (a[d] <= a[i]) {
              temp = 1;
              break;
            }
          }
          if (temp == 0) {
            pos = i + 1;
            break;
          }
        }
      }
    }
    cout << pos << endl;
  }
}

```