Have you ever used the chat application QQ? Well, in a chat group of QQ, administrators can muzzle a user for days.

In Boboniu's chat group, there's a person called Du Yi who likes to make fun of Boboniu every day.

Du will chat in the group for n days. On the i-th day:

  * If Du can speak, he'll make fun of Boboniu with fun factor a_i. But after that, he may be muzzled depending on Boboniu's mood. 
  * Otherwise, Du won't do anything. 



Boboniu's mood is a constant m. On the i-th day:

  * If Du can speak and a_i>m, then Boboniu will be angry and muzzle him for d days, which means that Du won't be able to speak on the i+1, i+2, ⋅⋅⋅, min(i+d,n)-th days. 
  * Otherwise, Boboniu won't do anything. 



The total fun factor is the sum of the fun factors on the days when Du can speak.

Du asked you to find the maximum total fun factor among all possible permutations of a.

Input

The first line contains three integers n, d and m (1≤ d≤ n≤ 10^5,0≤ m≤ 10^9).

The next line contains n integers a_1, a_2, …,a_n (0≤ a_i≤ 10^9).

Output

Print one integer: the maximum total fun factor among all permutations of a.

Examples

Input


5 2 11
8 10 15 23 5


Output


48


Input


20 2 16
20 5 8 2 18 16 2 16 16 1 5 16 2 13 6 16 4 17 21 7


Output


195

Note

In the first example, you can set a'=[15, 5, 8, 10, 23]. Then Du's chatting record will be:

  1. Make fun of Boboniu with fun factor 15. 
  2. Be muzzled. 
  3. Be muzzled. 
  4. Make fun of Boboniu with fun factor 10. 
  5. Make fun of Boboniu with fun factor 23. 



Thus the total fun factor is 48.
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a < 0) a = -a;
  if (b < 0) b = -b;
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
void read(T &res) {
  bool flag = false;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
  for (res = ch - 48; isdigit(ch = getchar());
       res = (res << 1) + (res << 3) + ch - 48)
    ;
  flag && (res = -res);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long qp(long long a, long long b, long long mod) {
  long long ans = 1;
  if (b == 0) {
    return ans % mod;
  }
  while (b) {
    if (b % 2 == 1) {
      b--;
      ans = ans * a % mod;
    }
    a = a * a % mod;
    b = b / 2;
  }
  return ans % mod;
}
long long qpn(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % p;
      --b;
    }
    a = (a * a) % p;
    b >>= 1;
  }
  return ans % p;
}
long long fact_pow(long long n, long long p) {
  long long res = 0;
  while (n) {
    n /= p;
    res += n;
  }
  return res;
}
long long mult(long long a, long long b, long long p) {
  a %= p;
  b %= p;
  long long r = 0, v = a;
  while (b) {
    if (b & 1) {
      r = (r + v) % p;
      r = (r + p) % p;
    }
    v <<= 1;
    v = (v + p) % p;
    b >>= 1;
  }
  return r % p;
}
long long pow_mod(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = mult(res, x, mod);
    x = mult(x, x, mod);
    n >>= 1;
  }
  return res;
}
long long quick_pow(long long a, long long b, long long p) {
  long long r = 1, v = a % p;
  while (b) {
    if (b & 1) r = mult(r, v, p);
    v = mult(v, v, p);
    b >>= 1;
  }
  return r;
}
bool CH(long long a, long long n, long long x, long long t) {
  long long r = quick_pow(a, x, n);
  long long z = r;
  for (long long i = 1; i <= t; i++) {
    r = mult(r, r, n);
    if (r == 1 && z != 1 && z != n - 1) return true;
    z = r;
  }
  return r != 1;
}
bool Miller_Rabin(long long n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (!(n & 1)) return false;
  long long x = n - 1, t = 0;
  while (!(x & 1)) {
    x >>= 1;
    t++;
  }
  srand(time(NULL));
  long long o = 8;
  for (long long i = 0; i < o; i++) {
    long long a = rand() % (n - 1) + 1;
    if (CH(a, n, x, t)) return false;
  }
  return true;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long ans = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return ans;
}
long long INV(long long a, long long b) {
  long long x, y;
  return exgcd(a, b, x, y), (x % b + b) % b;
}
long long crt(long long x, long long p, long long mod) {
  return INV(p / mod, mod) * (p / mod) * x;
}
long long FAC(long long x, long long a, long long b) {
  if (!x) return 1;
  long long ans = 1;
  for (long long i = 1; i <= b; i++)
    if (i % a) ans *= i, ans %= b;
  ans = pow_mod(ans, x / b, b);
  for (long long i = 1; i <= x % b; i++)
    if (i % a) ans *= i, ans %= b;
  return ans * FAC(x / a, a, b) % b;
}
long long C(long long n, long long m, long long a, long long b) {
  long long N = FAC(n, a, b), M = FAC(m, a, b), Z = FAC(n - m, a, b), sum = 0,
            i;
  for (i = n; i; i = i / a) sum += i / a;
  for (i = m; i; i = i / a) sum -= i / a;
  for (i = n - m; i; i = i / a) sum -= i / a;
  return N * pow_mod(a, sum, b) % b * INV(M, b) % b * INV(Z, b) % b;
}
long long exlucas(long long n, long long m, long long p) {
  long long t = p, ans = 0, i;
  for (i = 2; i * i <= p; i++) {
    long long k = 1;
    while (t % i == 0) {
      k *= i, t /= i;
    }
    ans += crt(C(n, m, i, k), p, k), ans %= p;
  }
  if (t > 1) ans += crt(C(n, m, t, t), p, t), ans %= p;
  return ans % p;
}
long long H(long long x, long long p) {
  long long ans = 0;
  if (x == 0) return 1;
  x = x % (2 * p);
  if (x == 0) x = 2 * p;
  for (long long i = 2; i <= x; ++i)
    ans = (ans * i + (i % 2 == 0 ? 1 : -1)) % p;
  return (ans + p) % p;
}
long long d, n, m;
long long c1, c2, a[100005], b[100005], sum[100005];
signed main() {
  read(n);
  read(d);
  read(m);
  d++;
  for (long long i = 1; i <= n; i++) {
    long long x;
    read(x);
    if (x > m)
      b[++c2] = x;
    else
      a[++c1] = x;
  }
  sort(a + 1, a + c1 + 1, greater<long long>());
  sort(b + 1, b + c2 + 1, greater<long long>());
  for (long long i = 1; i <= c1; i++) {
    sum[i] = sum[i - 1] + a[i];
  }
  long long res = sum[c1], now = 0, day = n - 1 + d;
  for (long long i = 1; i <= c2; i++) {
    {
      now += b[i];
      day -= d;
      if (day < 0) break;
      res = max(res, now + sum[min(c1, day)]);
    }
  }
  cout << res;
  return 0;
}
