// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

typedef pair<int, int> P;

typedef long long ll;



const int maxv = 10000 + 10;

struct edge

{

    int to, d, c;

    edge() {}

    edge(int to, int d, int c):to(to), d(d), c(c) {}

};

vector<edge> G[maxv];

int d[maxv];



int N, M;



void dijkstra(int s)

{

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

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

    d[s] = 0;

    que.push(P(0, s));

    while(!que.empty())

    {

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

        int v = p.second;

        if(p.first > d[v]) continue;

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

        {

            edge e = G[v][i];

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

            {

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

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

            }

        }

    }

}



int main()

{

    while(scanf("%d%d", &N, &M)==2 && N)

    {

        for(int i=0; i<N; i++)  G[i].clear();

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

        {

            int u, v, d, c;

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

            u--; v--;

            G[u].push_back(edge(v, d, c));

            G[v].push_back(edge(u, d, c));

        }

        dijkstra(0);

        int ans = 0;

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

        {

            int mincost = 0x3f3f3f3f;

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

            {

                edge e = G[i][j];

                if(d[e.to]+e.d==d[i] && e.c<mincost)

                    mincost = e.c;

            }

            ans += mincost;

        }

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

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.228485051198843x):
    #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;

    }