// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 1.1540476873210928x):
#include <iostream>

#include <cstring>

const int SIZE = 10010;

const int MAX = 0x3f3f3f3f;



typedef struct Road

{

	int from;

	int to;

	int length;

	int cost;

	int next;

} Road;



int head[SIZE];

int dis[SIZE];

Road roads[SIZE * 5];

int front[SIZE];

bool visit[SIZE];



int city, road, capital;



void init()

{

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

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

	memset(visit, false, sizeof(visit));

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

	{

		dis[i] = MAX;

	}

	dis[1] = 0;

}



void dijkstra()

{

	while (true)

	{

		int pos = -1;

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

		{

			if (!visit[i] && (pos == -1 || dis[pos] > dis[i]))

			{

				pos = i;

			}

		}

		if (pos == -1)

		{

			break;

		}

		visit[pos] = true;

		for (int i = head[pos]; i != -1; i = roads[i].next)

		{

			int to = roads[i].to;

			if (dis[to] > dis[pos] + roads[i].length)

			{

				dis[to] = dis[pos] + roads[i].length;

				front[to] = i;

			}

			else if ((dis[to] == dis[pos] + roads[i].length) && front[to] != -1 && roads[i].cost < roads[front[to]].cost)

			{

				front[to] = i;

			}

		}

	}

}



int main()

{

	while (true)

	{

		std::cin >> city >> road;

		if (city == 0 && road == 0)

		{

			break;

		}

		init();

		int en = 0;

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

		{

			std::cin >> roads[en].from >> roads[en].to >> roads[en].length >> roads[en].cost;

			roads[en].next = head[roads[en].from];

			head[roads[en].from] = en;

			en++;

			roads[en].from = roads[en - 1].to;

			roads[en].to = roads[en - 1].from;

			roads[en].length = roads[en - 1].length;

			roads[en].cost = roads[en - 1].cost;

			roads[en].next = head[roads[en].from];

			head[roads[en].from] = en;

			en++;

		}

		dijkstra();

		int result = 0;

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

		{

			if (front[i] != -1)

			{

				result += roads[front[i]].cost;

			}

		}

		std::cout << result << std::endl;

	}

	return 0;

}