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

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <cstring>

#include <vector>

#include <cmath>

#include <queue>



using namespace std;

#define Fill(a, b) memset(a, b, sizeof(a))

#define Debug(x) cout<<#x<<"="<<(x)<<endl;

typedef pair<int, int> P;

const int inf = 0x1f1f1f1f;

const int max_n = 10010;

struct edge

{

    int from, to, cost, fee;

};

int n, ans;

//vector<int> g fee[max_n];

int d[max_n], pre[max_n];

vector<edge> G[max_n];

void dijkstra(int s);

vector<int> get_path(int t);

int main()

{

    int i = 0, j = 0, k = 0;

    int m, a, b, p, q;

    edge ei;

    while(scanf("%d %d", &n, &m)!=EOF)

    {

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

            break;

        ans = 0;

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

            G[i].clear();

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

        {

            scanf("%d %d %d %d", &a, &b, &p, &q);

            ei.from = a;

            ei.to = b;

            ei.cost = p;

            ei.fee = q;

            G[a].push_back(ei);

            ei.from = b;

            ei.to = a;

            G[b].push_back(ei);

        }

        dijkstra(1);



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

        /*

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

            cout << pre[i] << " ";

        cout << endl;

        */

    }



    return 0;

}

void dijkstra(int s)

{

    int i, j, u, v;

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

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

    {

        d[i] = inf;

    }

    Fill(pre, -1);

    d[s] = 0;

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

    while(!que.empty())

    {

        P pa = que.top();

        que.pop();

        v = pa.second;

        if(d[v] < pa.first)

            continue;

        edge es;

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

        {

            es = G[v][i];

            if(d[es.to] > d[es.from]+es.cost)

            {

                if(d[es.to]!=inf)

                    ans -= pre[es.to];

                d[es.to] = d[es.from]+es.cost;

                ans += es.fee;

                pre[es.to] = es.fee;

                que.push(P(d[es.to], es.to));

            }

            else if((d[es.to]==d[es.from]+es.cost)&&(pre[es.to]>es.fee))

            {

                ans = ans - pre[es.to] + es.fee;

                pre[es.to] = es.fee;

            }

        }

    }

}

// Faster C++ solution for the same problem (measured speedup: 1.2233273960135105x):
#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;

}