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

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <cstring>

#include <vector>

#include <cmath>

#include <queue>



using namespace std;

#define Fill(a, b) memset(a, b, sizeof(a))

#define Debug(x) cout<<#x<<"="<<(x)<<endl;

typedef pair<int, int> P;

const int inf = 0x1f1f1f1f;

const int max_n = 10010;

struct edge

{

    int from, to, cost, fee;

};

int n, ans;

//vector<int> g fee[max_n];

int d[max_n], pre[max_n];

vector<edge> G[max_n];

void dijkstra(int s);

vector<int> get_path(int t);

int main()

{

    int i = 0, j = 0, k = 0;

    int m, a, b, p, q;

    edge ei;

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

    {

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

            break;

        ans = 0;

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

            G[i].clear();

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

        {

            scanf("%d %d %d %d", &a, &b, &p, &q);

            ei.from = a;

            ei.to = b;

            ei.cost = p;

            ei.fee = q;

            G[a].push_back(ei);

            ei.from = b;

            ei.to = a;

            G[b].push_back(ei);

        }

        dijkstra(1);



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

        /*

        for(i = 2; i <= n; i++)

            cout << pre[i] << " ";

        cout << endl;

        */

    }



    return 0;

}

void dijkstra(int s)

{

    int i, j, u, v;

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

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

    {

        d[i] = inf;

    }

    Fill(pre, -1);

    d[s] = 0;

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

    while(!que.empty())

    {

        P pa = que.top();

        que.pop();

        v = pa.second;

        if(d[v] < pa.first)

            continue;

        edge es;

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

        {

            es = G[v][i];

            if(d[es.to] > d[es.from]+es.cost)

            {

                if(d[es.to]!=inf)

                    ans -= pre[es.to];

                d[es.to] = d[es.from]+es.cost;

                ans += es.fee;

                pre[es.to] = es.fee;

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

            }

            else if((d[es.to]==d[es.from]+es.cost)&&(pre[es.to]>es.fee))

            {

                ans = ans - pre[es.to] + es.fee;

                pre[es.to] = es.fee;

            }

        }

    }

}

// Faster C++ solution for the same problem (measured speedup: 1.2208895209676804x):
    #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;

    }