Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

The first line of input is the number of test case.
The first line of each test case contains three integers, N, M and R.
Then R lines followed, each contains three integers x_i, y_i and d_i.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ x_i < N
0 ≤ y_i < M
0 < d_i < 10000

#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define maxn 10000#define maxr 50005#define maxnr maxn+maxr+10int n, m, r, tot;        //1 ≤ N, M ≤ 10000struct edge{     int from, to, c;     edge(){        c = maxn;　　//初始化c=maxn的目的就是因为利用并查集时，我直接把男、女放在一个数组par中，    }　　　　　　　　　　//女的下标是从maxn+i开始的}cost[maxr * 2];bool cmp(edge a, edge b){ return a.c > b.c; }int par[maxr*2], k;        //并查集    int find(int x){        int r = x;    while (par[r] != r)r = par[r];    int i = x, j;    while (par[i] != r){        j = par[i];        par[i] = r;        i = j;    }    return r;}void unite(int x, int y){    x = find(x); y = find(y);    par[x] = par[y];}bool same(int x, int y){    x = find(x); y = find(y);    return x == y;}inline void init(){    for (int i = 0; i < maxr*2; i++)par[i] = i;    tot = 0;    k = 0;    int x, y, d;    for (int i = 0; i < r; i++){        scanf("%d %d %d", &x, &y, &d);        cost[++k].from = x; cost[k].to = maxn + y; cost[k].c = d;//        cost[++k].from = maxn + y; cost[k].to = x; cost[k].c = d;//look，从这里我就加了个偏移量maxn,并查集就可以用了    }    sort(cost + 1, cost + k + 1, cmp);}void solve(){for (int i = 1; i <= k; i++){        if (!same(cost[i].from, cost[i].to)){            tot += cost[i].c;            unite(cost[i].from, cost[i].to);        }    }    printf("%d\n", maxn*(n+m) - tot);}int main(){    int t;    scanf("%d", &t);    while (t--){        scanf("%d %d %d", &n, &m, &r);        init();        solve();    }//    system("pause");    return 0;}