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

#include <queue>

#include <functional>

#include <cstring>

using namespace std;



struct edge{

    int v, d, c;

    edge(){}

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

};



int N, M;

const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;//距?， ?点

int d[10665];



vector<edge> G[10665];



void dijkstra(int s){

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

    memset(d, INF, sizeof(d));

    d[s] = 0;

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

    while(!que.empty())

    {

        P p = que.top();

        que.pop();

        int q = p.second;

        if(d[q] < p.first) continue;

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

        {

            edge e = G[q][i];

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

            {

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

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

            }

        }

    }

}





int main()

{

    while(cin >> N >> M)

    {

        if(!N) break;

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

		{

			G[i].clear();

		}

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

        {

            int u,t,c,d;

            cin >> u >> t >> d >> c;

            u--;t--;

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

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

        }

        dijkstra(0);

        int ans=0;

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

        {

            int mincost = INF;

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

            {

                if(d[G[k][j].v]+G[k][j].d == d[k]

                 && G[k][j].c < mincost)

                    mincost = G[k][j].c;

            }

            ans += mincost;

        }

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.165856257037611x):
#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>



using namespace std;



const int inf = 0x3f3f3f3f;

const int N = 10414;

const int M = 40414;



struct Edge{

	int u, v, d, c;

	int next;

}edge[M];



int cnt, head[N];

int op, tp, link[N];

int vis[N], dis[N], cost[N];

queue<int>q;



void add_edge(int u, int v, int d, int c);

void spfa(int n);



int main(){

	int n, m, i, u, v, d, c;

	while(scanf("%d %d", &n, &m) && (n || m)){

		cnt = 0;

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

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

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

			add_edge(u, v, d, c);

			add_edge(v, u, d, c);

		}

		spfa(n);

	}

	return 0;

}

void add_edge(int u, int v, int d, int c){

	edge[cnt].u = u;

	edge[cnt].v = v;

	edge[cnt].d = d;

	edge[cnt].c = c;

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

	head[u] = cnt++;

}

void spfa(int n)

{

    int v, d, c;

    while(!q.empty())q.pop();

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

    memset(dis, inf, sizeof(dis));

    memset(cost, inf, sizeof(cost));

    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)

        {

            v = edge[i].v;

            d = edge[i].d;

            c = edge[i].c;

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

            {

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

                cost[v] = c;

                if(!vis[v])

                {

                    q.push(v);

                    vis[v]=true;

                }

            }

        }

    }

    int sum = 0;

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

    	sum += cost[i];

    }

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

    return;

}