// Original C++ solution (slower):
#include <iostream>

#include <stdio.h>

#include <string.h>

#include <queue>

using namespace std;

const int maxn=10050,maxm=20050,inf=0x3f3f3f3f;

struct Edge

{

    int to,next,dis,cos ;

}edge[maxm*2];

struct Pair

{

    int D,V,C;

    Pair(){;}

    Pair(int d,int c,int v)

    {

        D=d,V=v,C=c;

    }

    friend bool operator <(const Pair a,const Pair b)

    {

        if(a.D!=b.D)

        return a.D>b.D;

        else return a.C>b.C;

    }

};

int tot,head[maxn],dist[maxn],cost[maxn];

bool vis[maxn];

void init()

{

    tot=0;

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

}

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

{

    edge[tot].to=v;

    edge[tot].dis=d;

    edge[tot].cos=c;

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

    head[u]=tot++;

}

int dijstra(int start)

{

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

    memset(dist,inf,sizeof(dist));

    int ans=0;

    dist[start]=0;

   // vis[start]=true;

    priority_queue<Pair>que;

    que.push(Pair(0,0,start));

    while(!que.empty())

    {

        Pair now=que.top();

        que.pop();

        if(vis[now.V])

            continue;

        vis[now.V]=true;

        ans+=now.C;

        for(int i=head[now.V];i!=-1;i=edge[i].next)

        {

            int to=edge[i].to;

            que.push(Pair(now.D+edge[i].dis,edge[i].cos,to));

        }

    }

    return ans;

}

int main()

{

    int u,v,d,c;

    int n,m;

    while(scanf("%d%d",&n,&m)!=-1)

    {

        init();

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

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

        {

            scanf("%d%d%d%d",&u,&v,&d,&c);

            add_edge(u,v,d,c);

            add_edge(v,u,d,c);

        }

        printf("%d\n",dijstra(1));

    }

    return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>

#include <algorithm>



using namespace std;



constexpr int maxN = 10005;

constexpr int maxM = 400005;



int N, M;



int adj[maxN], gto[maxM], gnxt[maxM], length[maxM], cost[maxM], e;

void add_edge(int u, int v, int d, int c) {

    gto[e] = v; length[e] = d; cost[e] = c; gnxt[e] = adj[u]; adj[u] = e++;

}



void print_edge(int s) {

    printf("%d:", s);

    for (int e = adj[s]; ~e; e = gnxt[e])

        printf(" %d", gto[e]);

    putchar('\n');

}



int dis[maxN], pre[maxN];

int qu[maxN], qh, qt; bool inq[maxN];

void bellman_ford(int s) {

    memset(dis, 0x3f, sizeof dis); dis[s] = 0;

    for (inq[qu[qh = qt = 0] = s] = 1; qh <= qt; inq[qu[qh++ % N]] = 0)

        for (int u = qu[qh % N], e = adj[u], v = gto[e]; ~e; v = gto[e = gnxt[e]])

            if (dis[v] > dis[u] + length[e]) {

                dis[v] = dis[u] + length[e];

                pre[v] = cost[e];

                if (!inq[v]) inq[qu[++qt % N] = v] = 1;

            } else if (dis[v] == dis[u] + length[e]) {

                pre[v] = min(pre[v], cost[e]);

            }

}



int main() {

    while (~scanf("%d%d", &N, &M) && (N || M)) {

        memset(adj, -1, sizeof adj);

        for (int i = 0; i != M; ++i) {

            int u, v, d, c; scanf("%d%d%d%d", &u, &v, &d, &c);

            add_edge(u, v, d, c); add_edge(v, u, d, c);

        }

        bellman_ford(1);

        long long ans = 0;

        for (int i = 1; i <= N; ++i) ans += pre[i];

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

    }

    return 0;

}

// Fastest known solution for this problem: