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

#include<vector>

#include<queue>

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn  = 1e4+10;

const int INF = 0x3f3f3f3f;

typedef pair<int,int> P;

struct edge{

	int to,dis,cost;

		

};



int d[maxn];

int value[maxn];



int N,M;



int dijkstra(vector<edge>G[])

{



	memset(d,INF,sizeof(d));

	//memset(value,INF,sizeof(value));

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

		

	d[1] = 0;

	value[1] = 0;

	que.push(P(d[1],1));

	

	int res = 0;

	while(!que.empty())

	{

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

		

		int v = p.second;

		if(d[v] < p.first)

			continue;

			

		res += value[v];

		

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

		{

			edge e = G[v][i];

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

				value[e.to] = min(value[e.to],e.cost);

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

			{

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

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

				

				value[e.to] = e.cost;

			}

		}

	}

	return res;

	

}



int main()

{



	//freopen("c:\\users\\administrator\\desktop\\in.txt","r",stdin);

	while(cin >> N >> M, N||M)

	{

		

		vector<edge> G[maxn];

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

		{

			int u,v,d,c;

			edge e;

			cin >> u >> v >> d >> c;

			e.to= v, e.dis= d, e.cost =c;

			G[u].push_back(e);

			e.to= u;

			G[v].push_back(e);	

		}

		

		int res = dijkstra(G);

		cout << res <<endl;

	}		

} 

// Faster C++ solution for the same problem (measured speedup: 1.3493692139143179x):
#include <stdio.h>

#include <string.h>

#include <queue>

#include <iostream>

#include <algorithm>

using namespace std;

const int maxn=10050;

const int maxe=40414;

const int INF=0x3f3f3f3f;

struct note

{

    int v;

    int d;

    int c;

    int next;

};

note edge[maxe];

int head[maxn];

int cnt;

void init()

{

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

    cnt=0;

}

void addedge(int u,int v,int d,int c)

{

    edge[cnt].v=v;

    edge[cnt].c=c;

    edge[cnt].d=d;

    edge[cnt].next=head[u];

    head[u]=cnt++;

}

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 v=edge[i].v;

            int val=edge[i].d;

            if(dis[v]>dis[u]+val||(dis[v]==dis[u]+val&&cost[v]>edge[i].c))//有多条最短路时，取花费最小的

            {

                dis[v]=dis[u]+val;

                cost[v]=edge[i].c;

                if(!vis[v])

                {

                    q.push(v);

                    vis[v]=true;

                }

            }

        }

    }

}

int main()

{

    int n,m;

    int u,v,d,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,&d);

            addedge(u,v,c,d);

            addedge(v,u,c,d);

        }

        spfa(1,n);

        int ans=0;

        for(int i=1; i<=n; i++)

            ans+=cost[i];

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

    }

    return 0;

}