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



using namespace std;

const int N = 1E4 + 7;

bool vis[N];

int n, m;

typedef struct node

{

   int to,dis,cost;

   node(){}

   node(int a,int b,int c) : to(a),dis(b),cost(c) {}

   bool operator < (const node &a) const {

       return dis == a.dis ? cost > a.cost : dis > a.dis;

   }

}P;

vector<P>g[N];

int dij()

{

    int ans = 0;

    priority_queue<P> pq;

    pq.push(P(1,0,0));

    while(!pq.empty()) {

        P now = pq.top() ; pq.pop();

        if(vis[now.to]) continue;

        vis[now.to] = true;

        ans += now.cost;

        for(int i = 0;i < g[now.to].size();i ++) {

            P temp = g[now.to][i];

            pq.push(P(temp.to, now.dis + temp.dis, temp.cost));

        }

    }

    return ans;

}

int main()

{

   // freopen("f:\\jj.txt","r",stdin);

    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);

    while(cin >> n >> m) {

        if(n + m == 0)  break;

        for(int i = 1;i <= n;i ++)  g[i].clear();

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

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

            int a,b,c,d;

            cin >> a >> b >> c >> d;

            g[a].push_back(P(b,c,d));

            g[b].push_back(P(a,c,d));

        }

        cout << dij() << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2051491395207568x):
#include <stdio.h>

#include <string.h>

#include <queue>

#include <iostream>

#include <algorithm>

using namespace std;

const int maxn=10050;

const int maxe=40414;

const int INF=0x3f3f3f3f;

struct note

{

    int v;

    int d;

    int c;

    int next;

};

note edge[maxe];

int head[maxn];

int cnt;

void init()

{

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

    cnt=0;

}

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

{

    edge[cnt].v=v;

    edge[cnt].c=c;

    edge[cnt].d=d;

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

    head[u]=cnt++;

}

int cost[maxn];

int dis[maxn];

bool vis[maxn];

queue<int>q;

void spfa(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[1]=0;

    cost[1]=0;

    vis[1]=true;

    q.push(1);

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=false;

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

        {

            int v=edge[i].v;

            int val=edge[i].d;

            if(dis[v]>dis[u]+val||(dis[v]==dis[u]+val&&cost[v]>edge[i].c))//有多条最短路时，取花费最小的

            {

                dis[v]=dis[u]+val;

                cost[v]=edge[i].c;

                if(!vis[v])

                {

                    q.push(v);

                    vis[v]=true;

                }

            }

        }

    }

}

int main()

{

    int n,m;

    int u,v,d,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,&d);

            addedge(u,v,c,d);

            addedge(v,u,c,d);

        }

        spfa(n);

        int ans=0;

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

            ans+=cost[i];

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

    }

    return 0;

}