#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int OO = 0x3f3f3f3f, N = 2e5 + 7;
int a[N];
int n, x, y;
int main() {
  cin >> n >> x >> y;
  unordered_map<int, int> idx;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    idx[a[i]] = i;
  }
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    int c1 = 0, c2 = 0;
    for (int j = i - x; j < i; ++j) {
      if (j < 0 || a[j] > a[i]) {
        c1++;
      }
    }
    for (int j = i + 1; j <= i + y; ++j) {
      if (a[j] > a[i] || a[j] == 0) {
        c2++;
      }
    }
    if (c1 == x && c2 == y) {
      ans = i + 1;
      break;
    }
  }
  if (~ans) {
    cout << ans;
  } else {
    cout << n;
  }
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize("O3")
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int OO = 0x3f3f3f3f, N = 2e5 + 7;
int a[N];
int n, x, y;
int main() {
  cin >> n >> x >> y;
  unordered_map<int, int> idx;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    idx[a[i]] = i;
  }
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    int c1 = 0, c2 = 0;
    for (int j = i - x; j < i; ++j) {
      if (j < 0 || a[j] > a[i]) {
        c1++;
      }
    }
    for (int j = i + 1; j <= i + y; ++j) {
      if (a[j] > a[i] || a[j] == 0) {
        c2++;
      }
    }
    if (c1 == x && c2 == y) {
      ans = i + 1;
      break;
    }
  }
  if (~ans) {
    cout << ans;
  } else {
    cout << n;
  }
}

```