#include <bits/stdc++.h>
using namespace std;
void doin() {
  cin.tie();
  cout.tie();
  ios::sync_with_stdio(0);
}
template <typename X, typename Y>
istream& operator>>(istream& in, pair<X, Y>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
void getv(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv(vector<T>& v, int n, Ns... ns) {
  v.resize(n);
  for (auto& i : v) getv(i, ns...);
}
template <typename T>
void getv1(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv1(vector<T>& v, int n, Ns... ns) {
  v.resize(n + 1);
  for (int i = 1; i <= n; i++) getv1(v[i], ns...);
}
inline void getch(char& x) {
  while (x = getchar_unlocked(), x < 33) {
    ;
  }
}
inline void getstr(string& str) {
  str.clear();
  char cur;
  while (cur = getchar_unlocked(), cur < 33) {
    ;
  }
  while (cur > 32) {
    str += cur;
    cur = getchar_unlocked();
  }
}
template <typename T>
inline bool sc(T& num) {
  bool neg = 0;
  int c;
  num = 0;
  while (c = getchar_unlocked(), c < 33) {
    if (c == EOF) return false;
  }
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48;
  if (neg) num *= -1;
  return true;
}
template <typename a, typename b>
void minq(a& x, b y) {
  if (x > y) x = y;
}
template <typename a, typename b>
void maxq(a& x, b y) {
  if (x < y) x = y;
}
using AntonTsypko = void;
using arsijo = AntonTsypko;
using god = arsijo;
uniform_real_distribution<double> double_dist(0, 1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, x, y;
vector<long long> a;
bool chk(int i) {
  int l = max(0, i - x), r = min(i + y, (int)a.size() - 1);
  int m = INT_MAX;
  for (; l < i; l++) minq(m, a[l]);
  for (; r > i; r--) minq(m, a[r]);
  return m > a[i];
}
int main() {
  doin();
  cin >> n >> x >> y;
  getv(a, n);
  for (int i = 0; i < n; i++)
    if (chk(i)) return cout << i + 1, 0;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
void doin() {
  cin.tie();
  cout.tie();
  ios::sync_with_stdio(0);
}
template <typename X, typename Y>
istream& operator>>(istream& in, pair<X, Y>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
void getv(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv(vector<T>& v, int n, Ns... ns) {
  v.resize(n);
  for (auto& i : v) getv(i, ns...);
}
template <typename T>
void getv1(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv1(vector<T>& v, int n, Ns... ns) {
  v.resize(n + 1);
  for (int i = 1; i <= n; i++) getv1(v[i], ns...);
}
inline void getch(char& x) {
  while (x = getchar_unlocked(), x < 33) {
    ;
  }
}
inline void getstr(string& str) {
  str.clear();
  char cur;
  while (cur = getchar_unlocked(), cur < 33) {
    ;
  }
  while (cur > 32) {
    str += cur;
    cur = getchar_unlocked();
  }
}
template <typename T>
inline bool sc(T& num) {
  bool neg = 0;
  int c;
  num = 0;
  while (c = getchar_unlocked(), c < 33) {
    if (c == EOF) return false;
  }
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48;
  if (neg) num *= -1;
  return true;
}
template <typename a, typename b>
void minq(a& x, b y) {
  if (x > y) x = y;
}
template <typename a, typename b>
void maxq(a& x, b y) {
  if (x < y) x = y;
}
using AntonTsypko = void;
using arsijo = AntonTsypko;
using god = arsijo;
uniform_real_distribution<double> double_dist(0, 1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, x, y;
vector<long long> a;
bool chk(int i) {
  int l = max(0, i - x), r = min(i + y, (int)a.size() - 1);
  int m = INT_MAX;
  for (; l < i; l++) minq(m, a[l]);
  for (; r > i; r--) minq(m, a[r]);
  return m > a[i];
}
int main() {
  doin();
  cin >> n >> x >> y;
  getv(a, n);
  for (int i = 0; i < n; i++)
    if (chk(i)) return cout << i + 1, 0;
  return 0;
}

```