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

#include<iostream>

#include<algorithm>

#include<vector>

#include<queue>

#define MAX_V 10006

#define INF 0x3fffffff

using namespace std;

struct edge{

    int to,cost,dis;

    };

typedef pair<int,int> P;

int V;

vector<edge> G[MAX_V];

int d[MAX_V];



void dij(int s){

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

    fill(d,d+V,INF);

    d[s]=0;

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

    while(!que.empty()){

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

        int v=p.second;

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

        for(int 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;

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

            }

        }

    }

    int ans=0;



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

        int mimcost=INF;

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

            if(d[G[i][j].to]+G[i][j].cost==d[i]&&G[i][j].dis<mimcost){



                mimcost=G[i][j].dis;

                //cout<<mimcost<<endl;

            }

        }

        //cout<<mimcost<<endl;

        ans+=mimcost;

    }

    cout<<ans<<endl;



}



int main(){

   // freopen("1.txt","r",stdin);

    int M;

    while(cin>>V>>M){

        if(V==0&&M==0)break;

        for(int i=0;i<V;i++)G[i].clear();

        while(M--){

            int a,b,t,c;

            cin>>a>>b>>t>>c;

            a--;b--;

            edge tmp;

            tmp.to=b;

            tmp.cost=t;

            tmp.dis=c;

            G[a].push_back(tmp);

            tmp.to=a;

            G[b].push_back(tmp);

        }

        dij(0);

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1967186720727183x):
    #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()

    {

        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;

    }