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

#include <vector>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <iostream>

#include <algorithm>

const int INF=0x3f3f3f3f;

using namespace std;

struct edge{

    int to,cost,w;

};

int V,d[10005],dis[10005];

vector<edge> G[20005];

typedef pair<int,int> P;

void dijkstra(int s){

    int i,v;

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

	fill(d,d+V+1,INF);

	fill(dis,dis+V+1,INF);

	d[s]=dis[s]=0;

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

	while(!que.empty()){

		P p=que.top(); que.pop();

		v=p.second;

		if(d[v]<p.first)

        continue;

		for(i=0;i<G[v].size();i++){

			edge e=G[v][i];

			if(d[e.to]>d[v]+e.cost){

				d[e.to]=d[v]+e.cost;

                dis[e.to]=e.w;

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

			}

			else if(d[e.to]==d[v]+e.cost)

            dis[e.to]=min(dis[e.to],e.w);

		}

	}

}

int main(){

    int m,st,en,i,j,x,y,z,q,ans;

    edge temp;

    while(scanf("%d%d",&V,&m)!=EOF&&(V||m)){

        for(i=0;i<=V;i++)

        G[i].clear();

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

            scanf("%d%d%d%d",&x,&y,&z,&q);

            temp.to=y;temp.cost=z;temp.w=q;

            G[x].push_back(temp);

            temp.to=x;

            G[y].push_back(temp);

        }

        dijkstra(1);

        ans=0;

        for(i=1;i<=V;i++)

        ans+=dis[i];

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

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4088681795779934x):
    #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;  

    }