// Slower C++ solution:
#include <iostream>

#include <queue>

#include <functional>

#include <cstring>

using namespace std;



struct edge{

    int v, d, c;

    edge(){}

    edge(int v, int d, int c) : v(v),d(d),c(c){}

};



int N, M;

const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;//距?， ?点

int d[10665];



vector<edge> G[10665];



void dijkstra(int s){

    priority_queue<P, vector<P>,greater<P> > que;

    memset(d, INF, sizeof(d));

    d[s] = 0;

    que.push(P(0,s));

    while(!que.empty())

    {

        P p = que.top();

        que.pop();

        int q = p.second;

        if(d[q] < p.first) continue;

        for(int i = 0; i < G[q].size(); i++)

        {

            edge e = G[q][i];

            if(d[e.v] > d[q] + e.d)

            {

                d[e.v] = d[q] + e.d;

                que.push(P(d[e.v], e.v));

            }

        }

    }

}





int main()

{

    while(cin >> N >> M)

    {

        if(!N) break;

        for(int i = 0; i < N; i++)

		{

			G[i].clear();

		}

        for(int i = 0; i < M; i++)

        {

            int u,t,c,d;

            cin >> u >> t >> d >> c;

            u--;t--;

            G[u].push_back(edge(t,d,c));

            G[t].push_back(edge(u,d,c));

        }

        dijkstra(0);

        int ans=0;

        for(int k = 1; k < N; k++)

        {

            int mincost = INF;

            for(int j = 0; j < G[k].size(); j++)

            {

                if(d[G[k][j].v]+G[k][j].d == d[k]

                 && G[k][j].c < mincost)

                    mincost = G[k][j].c;

            }

            ans += mincost;

        }

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2631129154763738x):
    #include <stdio.h>  

    #include <string.h>  

    #include <queue>  

    #include <iostream>  

    #include <algorithm>  

    using namespace std;  

    const int maxn=10050;  

    const int maxe=2005000;  

    const int INF=1e9;  

    struct note  

    {  

        int to;  

        int w;  

        int c;  

        int next;  

    };  

    note edge[maxe];  

    int head[maxn];  

    int ip;  

    void init()  

    {  

        memset(head,-1,sizeof(head));  

        ip=0;  

    }  

    void addedge(int u,int v,int w,int c)  

    {  

        edge[ip].to=v;  

        edge[ip].c=c;  

        edge[ip].w=w;  

        edge[ip].next=head[u];  

        head[u]=ip++;  

    }  

    int cost[maxn];  

    int dis[maxn];  

    bool vis[maxn];  

    queue<int>q;  

    void spfa(int s,int n)  

    {  

        while(!q.empty())q.pop();  

        for(int i=1;i<=n;i++)  

            cost[i]=dis[i]=INF;  

        memset(vis,false,sizeof(vis));  

        dis[s]=0;  

        cost[s]=0;  

        vis[s]=true;  

        q.push(s);  

        while(!q.empty())  

        {  

            int u=q.front();  

            q.pop();  

            vis[u]=false;  

            for(int i=head[u];i!=-1;i=edge[i].next)  

            {  

                int to=edge[i].to;  

                int val=edge[i].w;  

                if(dis[to]>dis[u]+val||(dis[to]==dis[u]+val&&cost[to]>edge[i].c))//有多条最短路时，取花费最小的  

                {  

                    dis[to]=dis[u]+val;  

                    cost[to]=edge[i].c;  

                    if(!vis[to])  

                    {  

                        q.push(to);  

                        vis[to]=true;  

                    }  

                }  

            }  

        }  

    }  

    int main()  

    {  

       // freopen("cin.txt","r",stdin);  

        int n,m;  

        int u,v,w,c;  

        while(~scanf("%d%d",&n,&m))  

        {  

            if(n==0&&m==0) break;  

            init();  

            for(int i=0;i<m;i++)  

            {  

                scanf("%d%d%d%d",&u,&v,&c,&w);  

                addedge(u,v,c,w);  

                addedge(v,u,c,w);  

            }  

            spfa(1,n);  

            int ans=0;  

            for(int i=1;i<=n;i++)  

                ans+=cost[i];  

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

        }  

        return 0;  

    }