Codeforces Round #277.5 (Div. 2) B. BerSU Ball【贪心/双指针/每两个跳舞的人可以配对，并且他们两个的绝对值只差小于等于1，求最多匹配多少对】

B. BerSU Ball

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., an(1 ≤ ai ≤ 100), where ai 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, ..., bm(1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.

Output

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

Examples

input

4
1 4 6 2
5
5 1 5 7 9

output

3

input

4
1 2 3 4
4
10 11 12 13

output

0

input

5
1 1 1 1 1
3
1 2 3

output

2


【题意】：有n个boy，m个girl，每个人都有自己的舞蹈技术等级，现规定只有boy和girl的等级相差不大于1才能构成一对舞伴，在每个人都不重复的情况下，问最多能构成多少对？
【分析】：贪心。把两个数组都从小到大排序，再依次用当前最小的去跟对方比，若符合条件，则双方下标都++；若自己太低，则自己下标++，否则对方下标++。
【代码】：

#include<bits/stdc++.h>
using namespace std;
int n,m,cnt;
int a[1050],b[1050];
int main()
{
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i]; sort(a,a+n);
    cin>>m;
    for(int i=0;i<m;i++) cin>>b[i]; sort(b,b+m);
    
    for(int i=0,j=0;i<n&&j<m;)
    {
        if(abs(a[i]-b[j])<=1)
        {
            i++;
            j++;
            cnt++;
        }
        else if(a[i]>b[j])
        {
            j++;
        }
        else
        {
            i++;
        }
    }
    cout<<cnt<<endl;
    return 0;
}