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

#include <queue>

#include <functional>

#include <cstring>

using namespace std;



struct edge{

    int v, d, c;

    edge(){}

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

};



int N, M;

const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;//距?， ?点

int d[10665];



vector<edge> G[10665];



void dijkstra(int s){

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

    memset(d, INF, sizeof(d));

    d[s] = 0;

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

    while(!que.empty())

    {

        P p = que.top();

        que.pop();

        int q = p.second;

        if(d[q] < p.first) continue;

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

        {

            edge e = G[q][i];

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

            {

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

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

            }

        }

    }

}





int main()

{

    while(cin >> N >> M)

    {

        if(!N) break;

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

		{

			G[i].clear();

		}

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

        {

            int u,t,c,d;

            cin >> u >> t >> d >> c;

            u--;t--;

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

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

        }

        dijkstra(0);

        int ans=0;

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

        {

            int mincost = INF;

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

            {

                if(d[G[k][j].v]+G[k][j].d == d[k]

                 && G[k][j].c < mincost)

                    mincost = G[k][j].c;

            }

            ans += mincost;

        }

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1335899447903408x):
#include <cstdio>

#include <queue>

#include <vector>

#include <algorithm>



using namespace std;

const int INF = 0x3f3f3f3f;



const int maxn = 10240;



struct edge {

	int to, dis, cost;

	edge(int t, int d, int c) : to(t), dis(d), cost(c) {}

};





typedef pair<int, int> p;

priority_queue<p, vector<p>, greater<p> > q;

vector<edge> g[maxn];



int N, M;



void read_in() {

	//scanf("%d%d", &N, &M);

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

		g[i].clear();



	int from, to, dis, cost;

	for(int i = 1; i <= M; i++) {

		scanf("%d%d%d%d", &from, &to, &dis, &cost);

		g[from].push_back(edge(to, dis, cost));

		g[to].push_back(edge(from, dis, cost));

	}

}



int solve() {

	q.push(p(0, 1));



	const int INF = 0x7777777;

	int d[maxn];



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

		d[i] = INF;

	d[1] = 0;



	while( !q.empty() ) {

		p t = q.top();

		q.pop();



		int to = t.second, dis = t.first;

        if(d[to] < dis) continue;

		for(unsigned int i = 0; i < g[to].size(); i++) {

			edge To = g[to][i];

			if(d[To.to] > d[to] + To.dis)

				d[To.to] = d[to] + To.dis,

				q.push(p(d[To.to], To.to));

		}

	}

	//show(d);

    int ans = 0;



    for(int i = 2; i <= N; i++) {

        int tmp = INF;



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

            edge e = g[i][j];

            if(d[e.to] + e.dis == d[i] and e.cost < tmp)

                tmp = e.cost;

        }

        ans += tmp;

    }



    return ans;

}



int main() {

	while(scanf("%d %d", &N, &M) and N) {

		read_in();

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

	}

	return 0;

}