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

#include<queue>

#include<cstdlib>

#include<cstdio>

#include<cstring>



using namespace std;

const int maxn=4e4+100;

const int inf=0x3f3f3f3f;



struct Edge{

    int next;

    int to;

    int w;

    int  c;



}G[maxn];



int inq[maxn];

int dis[maxn];

int head[maxn];

int k=0;





void adde(int v,int to,int w,int c)//加?

{

    G[k].to=to;//第k条?指向的点

    G[k].w=w;//第k条?的??

    G[k].c=c;

    G[k].next=head[v];//与第k条??同起点的上一条?的?存位置

    head[v]=k++;

}



queue<int>Q;



void spfa(int s)

{

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

        memset(inq,0,sizeof(inq));

        memset(dis,inf,sizeof(dis));

        dis[s] = 0;

        inq[s] = 1;

        Q.push(s);

        while(!Q.empty())

        {

                int u = Q.front();

                Q.pop();

                inq[u]=0;

                for(int i=head[u];i!=-1;i=G[i].next)//遍?以u?起点的所有?

                {

                        int v=G[i].to;

                        if(dis[v]>dis[u]+G[i].w)

                        {

                                dis[v]=dis[u]+G[i].w;

                                if(inq[v]==0)

                                {

                                    Q.push(v);

                                    inq[v]=1;

                                }

                        }

                }

                inq[u]=0;



        }

}



int main()

{

        int n,m;

        while(scanf("%d%d",&n,&m)!=EOF)

        {

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

                        return 0;

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



                k=0;

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

                {

                            int a,b,c,d;

                            scanf("%d%d%d%d",&a,&b,&c,&d);

                            adde(a,b,c,d);

                            adde(b,a,c,d);

                }

                spfa(1);

                int ans=0;

                for(int i=2;i<=n;++i)

                {

                        int mi=inf;

                        for(int j=head[i];j!=-1;j=G[j].next)

                        {

                                if(dis[G[j].to]+G[j].w==dis[i]&&G[j].c<mi)

                                        mi=G[j].c;

                        }

                        ans+=mi;

                }

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

        }

        return 0;



}

// Faster C++ solution for the same problem (measured speedup: 1.5296241882474997x):
    #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;

    }