Codeforces Round #277.5(Div. 2) B. BerSU Ball【二分匹配】

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本文介绍了一种解决舞会中男女舞伴配对问题的算法。通过输入男女双方的舞蹈技能值，采用二分匹配的方法，计算出最大数量的配对组合。文章提供了完整的代码实现。

The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.

We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.

For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100), where a_i is the i-th boy's dancing skill.

Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b₁, b₂, ..., b_m (1 ≤ b_j ≤ 100), where b_j is the j-th girl's dancing skill.

Output

Print a single number — the required maximum possible number of pairs.

Examples

Input

Output

Input

4
1 2 3 4
4
10 11 12 13

Output

Input

Output

题目大意：

给你N个男生，以及M个女生，问你最多能够凑成几对一起跳舞。

每个人都有一个权值，只有异性之间才能凑成一对，并且两个人的能力值相差不超过1.

思路：

暴力建图跑二分匹配即可。

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
vector<int >mp[1150];
int a[1150];
int b[1150];
int match[1150];
int vis[1150];
int find(int u)
{
    for(int i=0;i<mp[u].size();i++)
    {
        int v=mp[u][i];
        if(vis[v]==0)
        {
            vis[v]=1;
            if(match[v]==-1||find(match[v]))
            {
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int n,m;
    while(~scanf("%d",&n))
    {
        memset(match,-1,sizeof(match));
        for(int i=1;i<=n;i++)mp[i].clear();
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=1;i<=m;i++)scanf("%d",&b[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(abs(a[i]-b[j])<=1)
                {
                    mp[i].push_back(j);
                }
            }
        }
        int output=0;
        for(int i=1;i<=n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(find(i))output++;
        }
        printf("%d\n",output);
    }
}