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

#include <cstdlib>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;

#define MAXN 10010

#define MAXM 40010



int n,m;

struct node

{

    int v,d,c,next;

}e[MAXM];

int ecnt,pre[MAXN],dis[MAXN],co[MAXN];

bool vis[MAXN];

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

{

    e[ecnt].v=v;

    e[ecnt].d=d;

    e[ecnt].c=c;

    e[ecnt].next=pre[u];

    pre[u]=ecnt++;

}

void spfa()

{

    queue<int> q;

    q.push(1);

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

    memset(dis,0x3f,sizeof(dis));

    memset(co,0x3f,sizeof(co));

    dis[1]=co[1]=0;

    while (!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=0;

        for (int i=pre[u];i!=-1;i=e[i].next)

        {

            int v=e[i].v;

            if ((dis[v]>dis[u]+e[i].d)||(dis[v]==dis[u]+e[i].d&&co[v]>e[i].c))

            {

                dis[v]=dis[u]+e[i].d;

                co[v]=e[i].c;

                if (!vis[v])

                {

                    vis[v]=1;

                    q.push(v);

                }

            }

        }

    }

}

int main()

{

    //freopen("/home/moor/Code/input","r",stdin);

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

    {

        ecnt=0;

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

        int u,v,d,c;

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

        {

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

            add(u,v,d,c);

            add(v,u,d,c);

        }

        spfa();

        int ans=0;

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

            ans+=co[i];

        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: