#include <bits/stdc++.h>
using namespace std;
long long int power_mod(long long int a, long long int x) {
  if (x == 0) return 1;
  long long int y = power_mod(a, x / 2);
  long long int ans = (y * y) % 1000000007;
  if (x % 2) ans = (ans * a) % 1000000007;
  return ans;
}
long long int inv(long long int a) { return power_mod(a, 1000000007 - 2); }
long long int power(long long int a, long long int x) {
  if (x == 0) return 1;
  long long int y = power(a, x / 2);
  long long int ans = (y * y);
  if (x % 2) ans *= a;
  return ans;
}
template <class T>
void printvec(vector<T> &a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
template <class T>
void printarr(T a[], int n) {
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, x, y;
  cin >> n >> x >> y;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int MAXN = n;
  int N = n;
  int K = 25;
  int log[MAXN + 1];
  log[1] = 0;
  for (int i = 2; i <= MAXN; i++) log[i] = log[i / 2] + 1;
  int st[MAXN][K];
  for (int i = 0; i < N; i++) st[i][0] = a[i];
  for (int j = 1; j <= K; j++)
    for (int i = 0; i + (1 << j) <= N; i++)
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  for (int i = 0; i < n; i++) {
    int L = i - x;
    L = max(L, 0);
    int R = i + y;
    R = min(R, n - 1);
    int j = log[R - L + 1];
    int minimum = min(st[L][j], st[R - (1 << j) + 1][j]);
    if (a[i] == minimum) {
      cout << i + 1 << "\n";
      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;
long long int power_mod(long long int a, long long int x) {
  if (x == 0) return 1;
  long long int y = power_mod(a, x / 2);
  long long int ans = (y * y) % 1000000007;
  if (x % 2) ans = (ans * a) % 1000000007;
  return ans;
}
long long int inv(long long int a) { return power_mod(a, 1000000007 - 2); }
long long int power(long long int a, long long int x) {
  if (x == 0) return 1;
  long long int y = power(a, x / 2);
  long long int ans = (y * y);
  if (x % 2) ans *= a;
  return ans;
}
template <class T>
void printvec(vector<T> &a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
template <class T>
void printarr(T a[], int n) {
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, x, y;
  cin >> n >> x >> y;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int MAXN = n;
  int N = n;
  int K = 25;
  int log[MAXN + 1];
  log[1] = 0;
  for (int i = 2; i <= MAXN; i++) log[i] = log[i / 2] + 1;
  int st[MAXN][K];
  for (int i = 0; i < N; i++) st[i][0] = a[i];
  for (int j = 1; j <= K; j++)
    for (int i = 0; i + (1 << j) <= N; i++)
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  for (int i = 0; i < n; i++) {
    int L = i - x;
    L = max(L, 0);
    int R = i + y;
    R = min(R, n - 1);
    int j = log[R - L + 1];
    int minimum = min(st[L][j], st[R - (1 << j) + 1][j]);
    if (a[i] == minimum) {
      cout << i + 1 << "\n";
      return 0;
    }
  }
}

```