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

using namespace std;

struct node{

    int next,d,c;

};

const int maxn=10005;

int n,m;

int d[maxn],c[maxn];

bool vis[maxn];

void init()

{

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

        d[i]=c[i]=1e9,vis[i]=0;

}

int main()

{

    while(~scanf("%d%d",&n,&m))

    {

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

        init();vector<node> v[maxn];

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

        {

            int x,y,d,c;

            scanf("%d%d%d%d",&x,&y,&d,&c);

            v[x].push_back(node{y,d,c});

            v[y].push_back(node{x,d,c});

        }

        queue<int> q;

        q.push(1);

        d[1]=c[1]=0;vis[1]=1;

        while(!q.empty())

        {

            int u=q.front();q.pop();vis[u]=0;

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

            {

                int nx=v[u][i].next;

                if(d[nx]==d[u]+v[u][i].d&&c[nx]>v[u][i].c){

                        c[nx]=v[u][i].c;

                }

                if(d[nx]>d[u]+v[u][i].d)

                {

                    d[nx]=d[u]+v[u][i].d;

                    c[nx]=v[u][i].c;

                    if(vis[nx]) continue;

                    vis[nx]=1;

                    q.push(nx);

                }

            }

        }

        int ans=0;

        for(int i=2;i<=n;i++) ans+=c[i];

        cout<<ans<<endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4047562573001655x):
    #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;

    }