17B. Hierarchy

B. Hierarchy

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employeeai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.

Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.

Input

The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.

Output

Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.

Examples

input

47 2 3 141 2 52 4 13 4 11 3 5

output

11

input

31 2 323 1 23 1 3

output

-1

Note

In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.

借鉴大神的做法

#include<cstdio>#include<algorithm>using namespace std;const int maxn=0x3f3f3f3f;int result(int n,int x[]){int sum=0,kase=1;for(int i=1;i<=n;++i){if(x[i]==maxn){sum-=maxn;if(!kase){return -1;}kase=0;}sum+=x[i];}return sum;}int main(){int n;while(~scanf("%d",&n)){int temp,x[1005]={0};for(int i=1;i<=n;++i){scanf("%d",&temp);x[i]=maxn;}int m,a,b,c;scanf("%d",&m);for(int i=0;i<m;++i){scanf("%d%d%d",&a,&b,&c);if(x[b]>c)//留下最少花费的 {x[b]=c;}}printf("%d\n",result(n,x));}return 0;}