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

#include <cstring>

#include <queue>

#define inf 0x3f3f3f3f

#define maxm 10005

#define maxn 40005



using namespace std;



int m,n;

long long edgenum,head[maxm],dis[maxm],num[maxm],ans;

bool vis[maxm];



struct node

{

    int from,to,len,cost,next;

}edge[maxn];



void init()

{

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

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

    memset(dis,inf,sizeof(dis));

    memset(num,inf,sizeof(num));

    edgenum=0;

}



void add(int u,int v,int len,int cost)

{

    node tp={u,v,len,cost,head[u]};

    edge[edgenum]=tp;

    head[u]=edgenum++;

}



long long SPFA(int st)

{

     queue<int> q;

     q.push(st);

     vis[st]=1;

     dis[st]=0;

     num[st]=0;

     ans=0;

     while(!q.empty())

     {

          int tmp=q.front();

          q.pop();

          vis[tmp]=0;

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

          {

                if((dis[edge[i].to]>dis[tmp]+edge[i].len)||((dis[edge[i].to]==dis[tmp]+edge[i].len)&&(num[edge[i].to]>edge[i].cost)))

                {

                    dis[edge[i].to]=dis[tmp]+edge[i].len;

                    num[edge[i].to]=edge[i].cost;

                    if(!vis[edge[i].to])

                    {

                        q.push(edge[i].to);

                        vis[edge[i].to]=1;

                    }

                }

          }

     }

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

        ans+=num[i];

     return ans;

}



int main()

{

    int u,v,len,cost;

    while(cin>>m>>n&&m)

    {

        init();

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

        {

            cin>>u>>v>>len>>cost;

            add(u,v,len,cost);

            add(v,u,len,cost);

        }

        cout<<SPFA(1)<<endl;

    }

    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: