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Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park.

She has only two pockets. She can put at mostk pebbles in each pocket at the same time. There aren different pebble types in the park, and there arew_i pebbles of thei-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day.

Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.

Input

The first line contains two integers n andk (1 ≤ n ≤ 10⁵,1 ≤ k ≤ 10⁹) — the number of different pebble types and number of pebbles Anastasia can place in one pocket.

The second line contains n integers w₁, w₂, ..., w_n (1 ≤ w_i ≤ 10⁴) — number of pebbles of each type.

Output

The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.

Example

Input

3 22 3 4

Output

3

Input

5 43 1 8 9 7

Output

5

Note

In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day.

Optimal sequence of actions in the second sample case:

• In the first day Anastasia collects 8 pebbles of the third type.
• In the second day she collects 8 pebbles of the fourth type.
• In the third day she collects 3 pebbles of the first type and1 pebble of the fourth type.
• In the fourth day she collects 7 pebbles of the fifth type.
• In the fifth day she collects 1 pebble of the second type.

题意：一个人在公园见鹅卵石，他有两袋子，每个袋子最多装k个，但是一个带字一次不能装不同堆的石头，给你多堆石头，问你最少要用多少天能捡完

si

思路：如果一堆小于k那么计半天，如果一堆大于k小于2*k记1天，如果一堆大于2*k，那么把他分为若干堆 ，一部分的个数为2*k,剩下的一堆为n%2*k，规则还是和前边一样，最后技术

计数，整天数+半天数，如果半天数为偶数那么计为  半天数/2  ,否则计为  半天数/2+1

ac代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespacestd;

int main(){
    int n,k,h; while(cin>>n>>k)
    {
        int p=0;
        int q=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&h);
            if(h<=k)
                p++;
            if(h>k&&h<=2*k)
                q++;
            if(h>2*k)
            {
                q+=h/(2*k);
                int d=h%(2*k);
                if(d<=k&&d>0)
                p++;
                if(d>k)
                    q++;

            }
        }
        if(p%2==0)
            cout<<q+p/2<<endl;
        else
            cout<<q+p/2+1<<endl;

    }
    return 0;
}