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

#include<cstdio>

#include<vector>

#include<cstring>

#include<iostream>

#include<algorithm>

#define RG register

#define LL long long

#define fre(a) freopen(a".in","r",stdin);//freopen(a".out","w",stdout);

using namespace std;

const int MAXN=50000;

int n,m,num;

int head[MAXN],to[MAXN],len[MAXN],cost[MAXN],Next[MAXN];

int dis[MAXN],pre[MAXN],pree[MAXN];

bool vis[MAXN],use[MAXN];

queue<int>que;

void add(int f,int t,int l,int c)

{

   Next[++num]=head[f];

   to[num]=t;

   len[num]=l;

   cost[num]=c;

   head[f]=num;

}

void SPFA()

{

   que.push(1);

   dis[1]=0;

   while(!que.empty())

      {

	 int now=que.front();

	 que.pop(); vis[now]=0;

	 for(int i=head[now];i;i=Next[i])

	    {

	       int son=to[i];

	       if(dis[now]+len[i]<dis[son])

		  {

		     dis[son]=dis[now]+len[i];

		     pre[son]=now;

		     pree[son]=i;

		     if(!vis[son])

			{

			   vis[son]=1;

			   que.push(son);

			}

		  }

	       else if(dis[now]+len[i]==dis[son])

		  {

		     if(cost[i]<cost[pree[son]])

			pree[son]=i,pre[son]=now;

		  }

	    }

      }

}

int main()

{

   //fre("1");

   while(1)

      {

	 scanf("%d%d",&n,&m);

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

	 num=0;

	 memset(head,0,sizeof head);

	 for(int i=1,a,b,c,d;i<=m;i++)

	    {

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

	       add(a,b,c,d); add(b,a,c,d);

	    }

	 memset(dis,0x3f3f3f3f,sizeof dis);

	 memset(vis,0,sizeof vis);

	 memset(use,0,sizeof use);

	 //memset(pree,0,sizeof pree);

	 SPFA();

	 //for(int i=1;i<=n;i++)printf("dis[%d]=%d\n",i,dis[i]);printf("----------------\n");

	 int ans=0;

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

	    for(int j=i;j>1;j=pre[j])

	       if(!use[pree[j]])

		  ans+=cost[pree[j]],use[pree[j]]=1;

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

      }

   return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.5521296987358841x):
    #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;

    }