#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
long long nop = 100;
vector<int> primes(nop, 1);
long long cnt = 0;
void sieve() {
  primes[1] = primes[0] = 0;
  for (long long i = 2; i * i <= nop; i++) {
    if (primes[i]) {
      for (long long j = i * i; j <= nop; j += i) {
        primes[j] = 0;
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (b > a) swap(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
void extgcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = a;
    y = b;
    return;
  }
  extgcd(b % a, a, x, y);
  long long y1 = y, x1 = x;
  x = y1 - (b / a) * x1;
  y = x1;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long bexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans *= num;
    }
    num *= num;
    e >>= 1;
  }
  return ans;
}
long long mexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans = (ans * num) % mod;
    }
    num = (num * num) % mod;
    e >>= 1;
  }
  return ans % mod;
}
long long modinv(long long a) { return bexp(a, mod - 2); }
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long n, x, y;
  cin >> n >> x >> y;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = -1;
  for (long long i = 0; i < n; i++) {
    int res = 1;
    for (long long j = 1; j <= x; j++) {
      if (i - j < 0) break;
      if (a[i] >= a[i - j]) {
        res = 0;
        break;
      }
    }
    for (long long j = 1; j <= y; j++) {
      if (i + j >= n) break;
      if (a[i] >= a[i + j]) {
        res = 0;
        break;
      }
    }
    if (res) {
      ans = i + 1;
      break;
    }
  }
  cout << ans << endl;
  return 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;
long long mod = 998244353;
long long nop = 100;
vector<int> primes(nop, 1);
long long cnt = 0;
void sieve() {
  primes[1] = primes[0] = 0;
  for (long long i = 2; i * i <= nop; i++) {
    if (primes[i]) {
      for (long long j = i * i; j <= nop; j += i) {
        primes[j] = 0;
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (b > a) swap(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
void extgcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = a;
    y = b;
    return;
  }
  extgcd(b % a, a, x, y);
  long long y1 = y, x1 = x;
  x = y1 - (b / a) * x1;
  y = x1;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long bexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans *= num;
    }
    num *= num;
    e >>= 1;
  }
  return ans;
}
long long mexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans = (ans * num) % mod;
    }
    num = (num * num) % mod;
    e >>= 1;
  }
  return ans % mod;
}
long long modinv(long long a) { return bexp(a, mod - 2); }
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long n, x, y;
  cin >> n >> x >> y;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = -1;
  for (long long i = 0; i < n; i++) {
    int res = 1;
    for (long long j = 1; j <= x; j++) {
      if (i - j < 0) break;
      if (a[i] >= a[i - j]) {
        res = 0;
        break;
      }
    }
    for (long long j = 1; j <= y; j++) {
      if (i + j >= n) break;
      if (a[i] >= a[i + j]) {
        res = 0;
        break;
      }
    }
    if (res) {
      ans = i + 1;
      break;
    }
  }
  cout << ans << endl;
  return 0;
}

```