/*
裸的最大权闭合图
解：参见胡波涛的《最小割模型在信息学竞赛中的应用
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
using namespace std;
#define N 55100//刚开始开的是5100一直越界应该是n+m
#define NN 510000
#define inf 0x3fffffff
struct node {
int u,v,w,next;
}bian[NN*8];
int head[N],yong,dis[N],work[N];
void init() {
yong=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w) {
bian[yong].v=v;
bian[yong].w=w;
bian[yong].next=head[u];
head[u]=yong++;
}
int bfs(int s,int t)
{
memset(dis,-1,sizeof(dis));
queue<int>q;
q.push(s);
dis[s]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u];i!=-1;i=bian[i].next)
{
int v=bian[i].v;
if(bian[i].w&&dis[v]==-1)
{
dis[v]=dis[u]+1;
q.push(v);
if(v==t)
return 1;
}
}
}
return 0;
}
int dfs(int s,int limit,int t)
{
if(s==t)return limit;
for(int &i=work[s];i!=-1;i=bian[i].next)
{
int v=bian[i].v;
if(bian[i].w&&dis[v]==dis[s]+1)
{
int tt=dfs(v,min(limit,bian[i].w),t);
if(tt)
{
bian[i].w-=tt;
bian[i^1].w+=tt;
return tt;
}
}
}
return 0;
}
int dinic(int s,int t)
{
int ans=0;
while(bfs(s,t))
{
memcpy(work,head,sizeof(head));
while(int tt=dfs(s,inf,t))
ans+=tt;
}
return ans;
}
int main(){
int n,m,i,k,sum,u,v,w,s,t;
while(scanf("%d%d",&n,&m)!=EOF) {
init();
s=0;t=n+m+1;
for(i=1;i<=n;i++) {
scanf("%d",&k);
addedge(i,t,k);
addedge(t,i,0);
}
sum=0;
for(i=1;i<=m;i++) {
scanf("%d%d%d",&u,&v,&w);
addedge(s,i+n,w);
addedge(i+n,s,0);
addedge(i+n,u,inf);
addedge(u,i+n,0);
addedge(i+n,v,inf);
addedge(v,i+n,0);
sum+=w;
}
printf("%d
",sum-dinic(s,t));
}
return 0;}