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

}

// 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: