Educational Codeforces Round 44 (Rated for Div. 2) C - Liebig's Barrels

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C. Liebig's Barrels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have m = n·k wooden staves. The i-th stave has length a_i. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.

Let volume v_j of barrel j be equal to the length of the minimal stave in it.

You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |v_x - v_y| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.

Input

The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 10⁵, 1 ≤ n·k ≤ 10⁵, 0 ≤ l ≤ 10⁹).

The second line contains m = n·k space-separated integers a₁, a₂, ..., a_m (1 ≤ a_i ≤ 10⁹) — lengths of staves.

Output

Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |v_x - v_y| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Examples

Input

Copy

4 2 1
2 2 1 2 3 2 2 3

Output

Copy

7

Input

Copy

2 1 0
10 10

Output

Copy

20

Input

Copy

1 2 1
5 2

Output

Copy

2

Input

Copy

3 2 1
1 2 3 4 5 6

Output

Copy

0

Note

In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].

In the second example you can form the following barrels: [10], [10].

In the third example you can form the following barrels: [2, 5].

In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.

题意：给你n*k个木板，每个木桶由k个木板组成，容量由最短的那块木板决定，并且要满足任意两个木桶之间的容量差不能大于l，否则的话输出0，反之输出n个木桶的最大容量之和。

分析：

贪心+模拟

mp1是一个栈，mp2是一个队列，ans记录每个木桶的容量之和。

先把n*k块木板按升序排列，把长度大于a[1]+l的放入mp2中，长度小于等于a[1]+l的木板放入mp1中，然后从mp1中依次取出一个木板，用mp2中的k-1个木板来凑一个木桶，当发现mp2中的木板不足以来凑时，那就用mp1中的木板来凑。最后输出ans就行了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
#include <vector>
#include <stack>
using namespace std;
typedef long long ll;
ll a[1000000];
int main()
{
    ll n,k,l;
    while(~scanf("%lld%lld%lld",&n,&k,&l))
    {
        stack<ll> mp1;    ///记录比a[1]+l小的数
        queue<ll> mp2;    ///记录比a[1]+l大的数
        for(int i=1; i<=n*k; i++)
            scanf("%lld",&a[i]);
        sort(a+1,a+1+n*k);
        ll maxn=a[1]+l;
        for(int i=1; i<=n*k; i++)
        {
            if(a[i]>maxn)
                mp2.push(a[i]);
            else
                mp1.push(a[i]);
        }
        ll ans=0;
        int cnt=0;
        if(mp1.size()<n)    ///满足题意的数字不足够拼成n个水桶
        {
            printf("0\n");
            continue;
        }
        else
        {
            while(!mp1.empty())
            {
                ll res=mp1.top();
                ans+=mp1.top();
                mp1.pop();
                int i,flog=0;
                for(i=1;i<k;i++)
                {
                    if(!mp2.empty())
                        mp2.pop();
                    else
                    {
                        flog=1;
                        break;
                    }
                }
                if(flog)
                {
                    for(;i<k;i++)
                    {
                        if(i==k-1)
                            ans+=mp1.top();
                        mp1.pop();
                    }
                    ans-=res;     ///初始时加上了但该木桶的容量不是res，所以要减去
                    break;
                }
            }
            while(!mp1.empty())
            {
                for(int j=1;j<=k;j++)
                {
                    if(j==k)
                        ans+=mp1.top();
                    mp1.pop();
                }
            }
            printf("%lld\n",ans);

        }
        while(!mp1.empty())
            mp1.pop();
        while(!mp2.empty())
            mp2.pop();
    }
    return 0;
}
/*
4 2 1
2 2 1 2 3 2 2 3
2 1 0
10 10
1 2 1
5 2
3 2 1
1 2 3 4 5 6
*/