// Original C++ solution (slower):
#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;

            }

        }

    }

}

// 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: