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

#include<cstdio>

#include<cstdlib>

#include<string>

#include<cstring>

#include<algorithm>

#include<queue>

#include<vector>

#define N 20100

#define M 50010

#define rg register

using namespace std;

typedef long long ll;

typedef double db;

int n,m,tot=0;

int dis[N],head[N],vis[N];

int nxt[M],to[M],w[M],c[M];

void link(int x,int y,int z,int v){

	nxt[tot]=head[x],to[tot]=y,w[tot]=z,c[tot]=v,head[x]=tot++;

}

void spfa(){

	queue <int> q;

	memset(dis,0x3f,sizeof(dis));

	memset(vis,0,sizeof(vis));

	q.push(1),dis[1]=0,vis[1]=1;

	while (!q.empty()){

		int x=q.front();q.pop();

		for (int i=head[x];i!=-1;i=nxt[i]){

			int v=to[i];

			if (dis[v]<=dis[x]+w[i]) continue;

			dis[v]=dis[x]+w[i];

			if (!vis[v]) vis[v]=1,q.push(v);

		}

		vis[x]=0;

	}

}

int main(){

//	freopen ("2249.in","r",stdin);

//	freopen ("2249.out","w",stdout);

	while (scanf ("%d%d",&n,&m)&&(n||m)){

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

		tot=0;int x,y,z,v;

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

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

			link(x,y,z,v),link(y,x,z,v);

		}

		spfa();

		int ans=0;

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

			int mx=2000;

			for (int j=head[i];j!=-1;j=nxt[j]){

				v=to[j];

				if (dis[i]==dis[v]+w[j]) mx=min(mx,c[j]);

			}

			ans+=mx;

		}

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

	

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.235584724127401x):
    #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;

    }