#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6, inf = 1e9, mod = 1e9 + 7;
long long tc, n, m, q, a, b, c, d, i, j, k, l, sum, ans, cnt1, cnt2, mx, mn,
    flag;
long long ar[N], cntl[N], cntr[N];
vector<long long> v;
string s;
int main() {
  cin >> n >> a >> b;
  for (i = 1; i <= n; i++) cin >> ar[i];
  for (i = 1; i <= n; i++) {
    for (j = i - 1, cnt1 = 0; cnt1 != a; j--)
      if (j == 0)
        cnt1 = a;
      else if (ar[j] > ar[i])
        cnt1++;
      else
        break;
    for (j = i + 1, cnt2 = 0; cnt2 != b; j++)
      if (j == n + 1)
        cnt2 = b;
      else if (ar[j] > ar[i])
        cnt2++;
      else
        break;
    if (cnt1 == a && cnt2 == b) cout << i, exit(0);
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long N = 1e6, inf = 1e9, mod = 1e9 + 7;
long long tc, n, m, q, a, b, c, d, i, j, k, l, sum, ans, cnt1, cnt2, mx, mn,
    flag;
long long ar[N], cntl[N], cntr[N];
vector<long long> v;
string s;
int main() {
  cin >> n >> a >> b;
  for (i = 1; i <= n; i++) cin >> ar[i];
  for (i = 1; i <= n; i++) {
    for (j = i - 1, cnt1 = 0; cnt1 != a; j--)
      if (j == 0)
        cnt1 = a;
      else if (ar[j] > ar[i])
        cnt1++;
      else
        break;
    for (j = i + 1, cnt2 = 0; cnt2 != b; j++)
      if (j == n + 1)
        cnt2 = b;
      else if (ar[j] > ar[i])
        cnt2++;
      else
        break;
    if (cnt1 == a && cnt2 == b) cout << i, exit(0);
  }
}

```