#include <bits/stdc++.h>

using namespace std;

#define N 500010

#define ll long long

#define INF (~0u>>1)

#define isdigit(c) ((c)>='0'&&(c)<='9') 

// https://www.luogu.com.cn/problem/P4174

inline int read(){

    ll x = , s = ;

    char c = getchar();

    while(!isdigit(c)){

        if(c == '-')s = -;

        c = getchar();

    }

    while(isdigit(c)){

        x = x *  + c - '';

        c = getchar();

    }

    return x * s;

}

struct node{

    int u, v, w;

    int next = -;

}t[N];

int f[N];

int ht, s;

int deth[N], cur[N];

int n, m;

int bian = -;//全部从-1开始

void addedge(ll u, ll v,ll w){

    bian++;

    t[bian].u = u;

    t[bian].v = v;

    t[bian].w = w;

    t[bian].next = f[u];

    f[u] = bian;

    return ;

}

inline void add(ll u, ll v,ll w){   //注意要加反边。这里就如此写了

    addedge(u, v, w);

    addedge(v, u, );

    return ;

}

queue <int> q;

bool bfs(int s, int ht){

    memset(deth, , sizeof(deth));

    while(!q.empty())q.pop();//进行初始化

    q.push(s);

    deth[s] = ;//起点记得设为-1

    while(!q.empty()){

        int now = q.front();q.pop();

        for(int i = f[now]; ~i; i = t[i].next){

            int v = t[i].v, u = t[i].u, w = t[i].w;

            if(!deth[v] && w > ){

                deth[v] = deth[u] + ; // 分层操作

                q.push(v);

            }

        }

    }

    return deth[ht] != ;

} 

ll dfs(int now,int dist){

    if(now == ht)return dist;

    for(int& i = cur[now]; ~i; i = t[i].next){ // 记得当前弧优化

        int w = t[i].w, v = t[i].v, u = t[i].u;

        if(deth[v] == deth[u] +  && w){

            int di = dfs(v, min(dist, w));

            if(di > ){

                t[i].w -= di;

                t[i^].w += di;

                return di;

            }

        }

    }

    return ;

}

ll Dicnic(){ // 经典Dicnic 操作，就是板子

    ll ans = ;

    while(bfs(s, ht)){

        memcpy(cur, f, sizeof(cur));

        while(ll temp = dfs(s,INF))

            ans += temp;

    }

    return ans;

}

int main(){

//    freopen("P4174_9.in", "r", stdin);

    memset(f, -, sizeof(f));

    n = read(), m = read();

    for(int i = ;i <= n; i++){

        int temp = read();

        add(, i, temp); // 从源点连向中转站

    }

    ll ans = ;

    int to1, to2, temp;

    for(int i = ;i <= m; i++){

        int to1 = read(), to2 = read(), temp = read();

        add(to1, i + n, INF);

        add(to2, i + n, INF);

        add(i + n,  + m + n, temp);//从用户连向汇点

        ans += temp;

    }

    s = , ht = n + m + ;

    ans -= Dicnic();//减掉届不到的用户和届到的中转站

    printf("%lld\n", ans);

    return ;

}