// 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: 1.235515494801036x):
#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 n)

{

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

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

        cost[i]=dis[i]=INF;

    memset(vis,false,sizeof(vis));

    dis[1]=0;

    cost[1]=0;

    vis[1]=true;

    q.push(1);

    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(n);

        int ans=0;

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

            ans+=cost[i];

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

    }

    return 0;

}