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

#include <algorithm>

#include <cstdio>

#include <queue>

#include <cstring>

#define INF 0x3f3f3f



using namespace std;

const int MAX_V=100001;

struct node

{

    int to, dis, cost, next;

}edge[MAX_V];



int head[MAX_V], d[MAX_V], mcost[MAX_V];//mcoat一个?点??一条?

bool vis[MAX_V];

int n, m, cnt;



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

{

    edge[cnt].to=v;

    edge[cnt].dis=d;

    edge[cnt].cost=c;

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

    head[u]=cnt++;

}



int spfa(int x)

{

    d[x]=0;         //必?！！！

    queue<int> que;

    que.push(x);

    vis[x]=true;

    while(!que.empty())

    {

        int u=que.front();

        que.pop();

        vis[u]=false;

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

        {

            node e = edge[i];

            if(d[e.to] > d[u]+e.dis)

            {

                mcost[e.to]=e.cost;     ////??最小花?

                d[e.to] = d[u]+e.dis;

                if(!vis[e.to])

                {

                    que.push(e.to);

                    vis[e.to]=true;

                }

            }

            else if(d[u]+e.dis==d[e.to] && e.cost<mcost[e.to])   ////??最小花?

                {

                    mcost[e.to]=e.cost;

                }

        }

    }

    int ans=0;

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

    {

     //   printf("\n%d\n", mcost[i]);

        ans+=mcost[i];

    }

    return ans;

}



int main()

{

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

    {

        cnt=0;          //从0?始！！！

        memset(head, -1, sizeof(head)); //初始化在前面！！！！

        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);      //双向的路！！！

        }



        //初始化

        memset(mcost, 0, sizeof(mcost));

        fill(d, d+n+1, INF);

        fill(vis, vis+n+1, false);



        int ans=spfa(1);

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

    }



    return 0;

}

// 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: