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

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

const int inf =0x3f3f3f3f;

const int MAXN=10005;

struct node

{

    int next, to, d, c;

} G[2000005];

int cur, n;

bool vis[MAXN];

int cost[MAXN];

int dist[MAXN];

int head[MAXN];

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

{

    G[cur].to=v;

    G[cur].c=c;

    G[cur].d=d;

    G[cur].next=head[u];

    head[u]=cur++;

}

queue<int>q;

void spfa(int s)

{



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

    memset(vis, false, sizeof(vis));

    fill(cost+1,cost+n+1,inf);

    fill(dist+1,dist+n+1,inf);

    q.push(s);

    dist[s]=0;

    cost[s]=0;

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=false;

        for(int i=head[u]; ~i; i=G[i].next)

        {

            int v=G[i].to;

            if(dist[v]>dist[u]+G[i].d||(dist[v]==dist[u]+G[i].d&&cost[v]>G[i].c))

            {

                dist[v]=dist[u]+G[i].d;

                cost[v]=G[i].c;

                if(!vis[v])

                {

                    vis[v]=true;

                    q.push(v);

                }

            }

        }

    }

}

int main()

{

    int m;

    int u, v, d, c;

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

    {

        if(n==0&&m==0)break;

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

        cur=0;

        while(m--)

        {

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

            add(u, v, d, c);

            add(v, u, d, c);

        }

        spfa(1);

        int ans=0;

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

            ans+=cost[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: