#include <bits/stdc++.h>
using namespace std;
long long int bg = 10000000000000000;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int power(long long int x, long long int y, long long int p) {
  if (x == 0) return 0;
  if (y == 0) return 1;
  long long int s = y / 2;
  long long int f = power(x, s, p) % p;
  if (y % 2 == 0)
    return (f * f) % p;
  else
    return (((f * f) % p) * x % p) % p;
}
void build1(long long int node, long long int seg[], long long int arr[],
            long long int start, long long int end) {
  if (start == end) {
    seg[node] = arr[start];
    return;
  }
  long long int mid = (start + end) / 2;
  build1(2 * node, seg, arr, start, mid);
  build1(2 * node + 1, seg, arr, mid + 1, end);
  seg[node] = seg[2 * node] + seg[2 * node + 1];
}
long long int query1(long long int node, long long int seg[],
                     long long int arr[], long long int start,
                     long long int end, long long int l, long long int r) {
  long long int mid = (start + end) / 2;
  if (r < start || l > end)
    return 0;
  else if (l <= start && r >= end)
    return seg[node];
  else
    return query1(2 * node, seg, arr, start, mid, l, r) +
           query1(2 * node + 1, seg, arr, mid + 1, end, l, r);
}
map<long long int, long long int> map1[25][25];
int main() {
  long long int i, j, n, m, k, x, y;
  cin >> n >> x >> y;
  long long int arr[n];
  for (i = 0; i < n; i++) cin >> arr[i];
  for (i = 0; i < n; i++) {
    long long int flag = 1;
    for (j = i + 1; j <= min(i + y, n - 1); j++)
      if (arr[j] <= arr[i]) flag = 0;
    for (j = i - 1; j >= max(i - x, n - 1); j--)
      if (arr[j] <= arr[i]) flag = 0;
    if (flag == 1) break;
  }
  cout << i + 1 << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 bg = 10000000000000000;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int power(long long int x, long long int y, long long int p) {
  if (x == 0) return 0;
  if (y == 0) return 1;
  long long int s = y / 2;
  long long int f = power(x, s, p) % p;
  if (y % 2 == 0)
    return (f * f) % p;
  else
    return (((f * f) % p) * x % p) % p;
}
void build1(long long int node, long long int seg[], long long int arr[],
            long long int start, long long int end) {
  if (start == end) {
    seg[node] = arr[start];
    return;
  }
  long long int mid = (start + end) / 2;
  build1(2 * node, seg, arr, start, mid);
  build1(2 * node + 1, seg, arr, mid + 1, end);
  seg[node] = seg[2 * node] + seg[2 * node + 1];
}
long long int query1(long long int node, long long int seg[],
                     long long int arr[], long long int start,
                     long long int end, long long int l, long long int r) {
  long long int mid = (start + end) / 2;
  if (r < start || l > end)
    return 0;
  else if (l <= start && r >= end)
    return seg[node];
  else
    return query1(2 * node, seg, arr, start, mid, l, r) +
           query1(2 * node + 1, seg, arr, mid + 1, end, l, r);
}
map<long long int, long long int> map1[25][25];
int main() {
  long long int i, j, n, m, k, x, y;
  cin >> n >> x >> y;
  long long int arr[n];
  for (i = 0; i < n; i++) cin >> arr[i];
  for (i = 0; i < n; i++) {
    long long int flag = 1;
    for (j = i + 1; j <= min(i + y, n - 1); j++)
      if (arr[j] <= arr[i]) flag = 0;
    for (j = i - 1; j >= max(i - x, n - 1); j--)
      if (arr[j] <= arr[i]) flag = 0;
    if (flag == 1) break;
  }
  cout << i + 1 << '\n';
  return 0;
}

```