CF981B Businessmen Problems map 模拟 二十二

Businessmen Problems

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies.

In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies.

All elements are enumerated with integers. The ChemForces company has discovered nn distinct chemical elements with indices a1,a2,…,ana1,a2,…,an, and will get an income of xixi Berland rubles if the ii-th element from this list is in the set of this company.

The TopChemist company discovered mm distinct chemical elements with indices b1,b2,…,bmb1,b2,…,bm, and it will get an income of yjyj Berland rubles for including the jj-th element from this list to its set.

In other words, the first company can present any subset of elements from {a1,a2,…,an}{a1,a2,…,an} (possibly empty subset), the second company can present any subset of elements from {b1,b2,…,bm}{b1,b2,…,bm} (possibly empty subset). There shouldn't be equal elements in the subsets.

Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105)  — the number of elements discovered by ChemForces.

The ii-th of the next nn lines contains two integers aiai and xixi (1≤ai≤1091≤ai≤109, 1≤xi≤1091≤xi≤109)  — the index of the ii-th element and the income of its usage on the exhibition. It is guaranteed that all aiai are distinct.

The next line contains a single integer mm (1≤m≤1051≤m≤105)  — the number of chemicals invented by TopChemist.

The jj-th of the next mm lines contains two integers bjbj and yjyj, (1≤bj≤1091≤bj≤109, 1≤yj≤1091≤yj≤109)  — the index of the jj-th element and the income of its usage on the exhibition. It is guaranteed that all bjbj are distinct.

Output

Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets.

Examples

input

Copy

3
1 2
7 2
3 10
4
1 4
2 4
3 4
4 4

output

Copy

24

input

Copy

1
1000000000 239
3
14 15
92 65
35 89

output

Copy

408

Note

In the first example ChemForces can choose the set (3,73,7), while TopChemist can choose (1,2,41,2,4). This way the total income is (10+2)+(4+4+4)=24(10+2)+(4+4+4)=24.

In the second example ChemForces can choose the only element 109109, while TopChemist can choose (14,92,3514,92,35). This way the total income is (239)+(15+65+89)=408(239)+(15+65+89)=408.

题意： 给你n个A公司的新元素及价值，再给你m个B公司的新元素及价值，两个公司的元素相同的取较大值，问两家公司元素价值总和

直接用一个map保存下相同元素的较大值，然后遍历相加就可以了

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + ;
const int mod = 1e9 + ;
typedef long long ll;
int main(){
    std::ios::sync_with_stdio(false);
    ll n, m;
    while( cin >> n ) {
        map<ll,ll> mm;
        ll x, y;
        for( ll i = ; i < n; i ++ ) {
            cin >> x >> y;
            if( mm[x] < y ) {
                mm[x] = y;
            }
        }
        cin >> m;
        for( ll i = ; i < m; i ++ ) {
            cin >> x >> y;
            if( mm[x] < y ) {
                mm[x] = y;
            }
        }
        ll sum = ;
        for( map<ll,ll>:: iterator it = mm.begin(); it != mm.end(); it ++ ) {
            sum += (*it).second;
        }
        cout << sum << endl;
    }
    return ;
}