// Original C++ solution (slower):
#include <iostream>

#include <cstring>

#include <cstdio>

#include <vector>

#include <queue>

#define maxn 10005

#define MAXN 10005

#define inf 0x0f0f0f0f



using namespace std;



int head[maxn],tol,dis[maxn],vis[maxn],n,m,mi[maxn];

int cost[MAXN];



struct node

{

    int a,len,to,next,cost;

}edge[50005];







void INIT()

{

    tol=0;

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

}



void add(int a,int b,int c,int d)

{

    edge[tol].to=b;

    edge[tol].len=c;

    edge[tol].cost=d;

    edge[tol].next=head[a];

    head[a]=tol++;

}



void SPFA()

{

    int i,a,p;



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

        dis[i]=inf,vis[i]=0,cost[i]=inf;

    queue<int> que;

    vis[0]=1,cost[i]=dis[0]=0;

    que.push(0);

    while(!que.empty())

    {

        a=que.front(),que.pop();

        vis[a]=0;

        for(p=head[a];p!=-1;p=edge[p].next)

        {

            if(dis[edge[p].to]>dis[a]+edge[p].len)

            {

                dis[edge[p].to]=dis[a]+edge[p].len;

                cost[edge[p].to]=edge[p].cost;

                if(!vis[edge[p].to])

                {

                    vis[edge[p].to]=1;

                    que.push(edge[p].to);

                }

            }

            if(dis[edge[p].to]==dis[a]+edge[p].len&&cost[edge[p].to]>edge[p].cost)

            {

            //    puts("Debug");

                cost[edge[p].to]=edge[p].cost;

                if(!vis[edge[p].to])

                {

                    vis[edge[p].to]=1;

                    que.push(edge[p].to);

                }

            }

        }

    }

}



int main(){

    int i,j;

    int a,b,l,c;

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

       INIT();

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

            scanf("%d%d%d%d",&a,&b,&l,&c);

            a--;

            b--;

            add(a,b,l,c);

            add(b,a,l,c);

       }

       SPFA();

       int res=0;

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

           // printf("Debug :%d\n",father[i]);

            res+=cost[i];

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

       //bfs();

    }

    return 0;

}

/*

5 6

1 2 2 2

2 3 1 1

1 4 1 1

3 5 1 1

4 2 2 1

4 5 1 1



*/

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>

#include <algorithm>



using namespace std;



constexpr int maxN = 10005;

constexpr int maxM = 400005;



int N, M;



int adj[maxN], gto[maxM], gnxt[maxM], length[maxM], cost[maxM], e;

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

    gto[e] = v; length[e] = d; cost[e] = c; gnxt[e] = adj[u]; adj[u] = e++;

}



void print_edge(int s) {

    printf("%d:", s);

    for (int e = adj[s]; ~e; e = gnxt[e])

        printf(" %d", gto[e]);

    putchar('\n');

}



int dis[maxN], pre[maxN];

int qu[maxN], qh, qt; bool inq[maxN];

void bellman_ford(int s) {

    memset(dis, 0x3f, sizeof dis); dis[s] = 0;

    for (inq[qu[qh = qt = 0] = s] = 1; qh <= qt; inq[qu[qh++ % N]] = 0)

        for (int u = qu[qh % N], e = adj[u], v = gto[e]; ~e; v = gto[e = gnxt[e]])

            if (dis[v] > dis[u] + length[e]) {

                dis[v] = dis[u] + length[e];

                pre[v] = cost[e];

                if (!inq[v]) inq[qu[++qt % N] = v] = 1;

            } else if (dis[v] == dis[u] + length[e]) {

                pre[v] = min(pre[v], cost[e]);

            }

}



int main() {

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

        memset(adj, -1, sizeof adj);

        for (int i = 0; i != M; ++i) {

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

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

        }

        bellman_ford(1);

        long long ans = 0;

        for (int i = 1; i <= N; ++i) ans += pre[i];

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

    }

    return 0;

}

// Fastest known solution for this problem: