#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, inf = 0x3f3f3f3f;
int a[maxn], x, y, n, l[maxn], r[maxn];
void solve() {
  scanf("%d %d", &x, &y);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  stack<int> st;
  for (int i = 1; i <= n; ++i) {
    if (st.empty() || a[i] > a[st.top()]) {
      st.push(i);
      l[i] = i - 1;
    } else {
      while (!st.empty() && a[i] < a[st.top()]) {
        st.pop();
      }
      if (st.empty())
        l[i] = 0;
      else
        l[i] = st.top();
      st.push(i);
    }
  }
  while (!st.empty()) st.pop();
  for (int i = n; i >= 1; --i) {
    if (st.empty() || a[i] > a[st.top()]) {
      st.push(i);
      r[i] = i + 1;
    } else {
      while (!st.empty() && a[i] < a[st.top()]) {
        st.pop();
      }
      if (st.empty())
        r[i] = n + 1;
      else
        r[i] = st.top();
      st.push(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (l[i] == 0) l[i] = -10;
    if (r[i] == n + 1) r[i] = n + 10;
    int tmp1 = i - l[i] - 1, tmp2 = r[i] - i - 1;
    if (tmp1 >= x && tmp2 >= y) {
      printf("%d\n", i);
      return;
    }
  }
  printf("no anwser!\n");
  return;
}
int main() {
  while (scanf("%d", &n) != EOF) solve();
  return 0;
}

### Prompt

Your task is to create 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;
const int maxn = 1e5 + 10, inf = 0x3f3f3f3f;
int a[maxn], x, y, n, l[maxn], r[maxn];
void solve() {
  scanf("%d %d", &x, &y);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  stack<int> st;
  for (int i = 1; i <= n; ++i) {
    if (st.empty() || a[i] > a[st.top()]) {
      st.push(i);
      l[i] = i - 1;
    } else {
      while (!st.empty() && a[i] < a[st.top()]) {
        st.pop();
      }
      if (st.empty())
        l[i] = 0;
      else
        l[i] = st.top();
      st.push(i);
    }
  }
  while (!st.empty()) st.pop();
  for (int i = n; i >= 1; --i) {
    if (st.empty() || a[i] > a[st.top()]) {
      st.push(i);
      r[i] = i + 1;
    } else {
      while (!st.empty() && a[i] < a[st.top()]) {
        st.pop();
      }
      if (st.empty())
        r[i] = n + 1;
      else
        r[i] = st.top();
      st.push(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (l[i] == 0) l[i] = -10;
    if (r[i] == n + 1) r[i] = n + 10;
    int tmp1 = i - l[i] - 1, tmp2 = r[i] - i - 1;
    if (tmp1 >= x && tmp2 >= y) {
      printf("%d\n", i);
      return;
    }
  }
  printf("no anwser!\n");
  return;
}
int main() {
  while (scanf("%d", &n) != EOF) solve();
  return 0;
}

```