#include <bits/stdc++.h>
using namespace std;
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
bool prime[100005];
void SieveOfEratosthenes(long long int n) {
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
bool isPrime(long long int n) {
  if (n < 2) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
std::vector<long long int> factors(long long int n) {
  std::vector<long long int> f;
  for (long long int i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      f.push_back(i);
      n = n / i;
    }
  }
  if (n > 1) f.push_back(n);
  return f;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int modpow(long long int x, long long int n, long long int m) {
  if (n == 0) return 1 % m;
  long long int u = modpow(x, n / 2, m);
  u = (u % m * u % m) % m;
  if (n % 2 == 1) u = (u % m * x % m) % m;
  return u % m;
}
long long int eulerstotientfunction(long long int n) {
  long long int ans = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n = n / i;
      ans -= ans / i;
    }
  }
  if (n > 1) ans -= ans / n;
  return ans;
}
long long int modinverse(long long int x, long long int m) {
  return modpow(x, m - 2, m);
}
long long int ncr(long long int n, long long int k) {
  long long int res = 1;
  k = min(k, n - k);
  for (long long int i = 0; i < k; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
int main() {
  fastio();
  long long int n, x, y;
  cin >> n >> x >> y;
  std::vector<long long int> arr(n);
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int flag = 0, i = 0;
  for (i = 0; i < n; i++) {
    flag = 0;
    for (long long int j = i - x; j < i && j >= 0 && j < n; j++) {
      if (arr[i] >= arr[j]) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) continue;
    for (long long int j = i + 1; j <= i + y && j < n; j++) {
      if (arr[i] >= arr[j]) {
        flag = 1;
        break;
      }
    }
    if (flag == 0) break;
  }
  cout << i + 1 << '\n';
}

### 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;
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
bool prime[100005];
void SieveOfEratosthenes(long long int n) {
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
bool isPrime(long long int n) {
  if (n < 2) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
std::vector<long long int> factors(long long int n) {
  std::vector<long long int> f;
  for (long long int i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      f.push_back(i);
      n = n / i;
    }
  }
  if (n > 1) f.push_back(n);
  return f;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int modpow(long long int x, long long int n, long long int m) {
  if (n == 0) return 1 % m;
  long long int u = modpow(x, n / 2, m);
  u = (u % m * u % m) % m;
  if (n % 2 == 1) u = (u % m * x % m) % m;
  return u % m;
}
long long int eulerstotientfunction(long long int n) {
  long long int ans = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n = n / i;
      ans -= ans / i;
    }
  }
  if (n > 1) ans -= ans / n;
  return ans;
}
long long int modinverse(long long int x, long long int m) {
  return modpow(x, m - 2, m);
}
long long int ncr(long long int n, long long int k) {
  long long int res = 1;
  k = min(k, n - k);
  for (long long int i = 0; i < k; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
int main() {
  fastio();
  long long int n, x, y;
  cin >> n >> x >> y;
  std::vector<long long int> arr(n);
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int flag = 0, i = 0;
  for (i = 0; i < n; i++) {
    flag = 0;
    for (long long int j = i - x; j < i && j >= 0 && j < n; j++) {
      if (arr[i] >= arr[j]) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) continue;
    for (long long int j = i + 1; j <= i + y && j < n; j++) {
      if (arr[i] >= arr[j]) {
        flag = 1;
        break;
      }
    }
    if (flag == 0) break;
  }
  cout << i + 1 << '\n';
}

```