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

#include <queue>

#include <vector>

#include <algorithm>



using namespace std;

const int INF = 0x3f3f3f3f;



const int maxn = 10240;



struct edge {

	int to, dis, cost;

	edge(int t, int d, int c) : to(t), dis(d), cost(c) {}

};





typedef pair<int, int> p;

priority_queue<p, vector<p>, greater<p> > q;

vector<edge> g[maxn];



int N, M;



void read_in() {

	//scanf("%d%d", &N, &M);

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

		g[i].clear();



	int from, to, dis, cost;

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

		scanf("%d%d%d%d", &from, &to, &dis, &cost);

		g[from].push_back(edge(to, dis, cost));

		g[to].push_back(edge(from, dis, cost));

	}

}



int solve() {

	q.push(p(0, 1));



	const int INF = 0x7777777;

	int d[maxn];



	for(int i = 2; i <= N; i++) 

		d[i] = INF;

	d[1] = 0;



	while( !q.empty() ) {

		p t = q.top();

		q.pop();



		int to = t.second, dis = t.first;

        if(d[to] < dis) continue;

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

			edge To = g[to][i];

			if(d[To.to] > d[to] + To.dis)

				d[To.to] = d[to] + To.dis,

				q.push(p(d[To.to], To.to));

		}

	}

	//show(d);

    int ans = 0;



    for(int i = 2; i <= N; i++) {

        int tmp = INF;



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

            edge e = g[i][j];

            if(d[e.to] + e.dis == d[i] and e.cost < tmp)

                tmp = e.cost;

        }

        ans += tmp;

    }



    return ans;

}



int main() {

	while(scanf("%d %d", &N, &M) and N) {

		read_in();

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

	}

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.1141966718749177x):
#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;

}