// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;

typedef pair<int, int> P;

typedef long long ll;



const int maxv = 10000 + 10;

struct edge

{

    int to, d, c;

    edge() {}

    edge(int to, int d, int c):to(to), d(d), c(c) {}

};

vector<edge> G[maxv];

int d[maxv];



int N, M;



void dijkstra(int s)

{

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

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

    d[s] = 0;

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

    while(!que.empty())

    {

        P p = que.top(); que.pop();

        int v = p.second;

        if(p.first > d[v]) continue;

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

        {

            edge e = G[v][i];

            if(d[e.to] > d[v]+e.d)

            {

                d[e.to] = d[v] + e.d;

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

            }

        }

    }

}



int main()

{

    while(scanf("%d%d", &N, &M)==2 && N)

    {

        for(int i=0; i<N; i++)  G[i].clear();

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

        {

            int u, v, d, c;

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

            u--; v--;

            G[u].push_back(edge(v, d, c));

            G[v].push_back(edge(u, d, c));

        }

        dijkstra(0);

        int ans = 0;

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

        {

            int mincost = 0x3f3f3f3f;

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

            {

                edge e = G[i][j];

                if(d[e.to]+e.d==d[i] && e.c<mincost)

                    mincost = e.c;

            }

            ans += mincost;

        }

        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: