// Original C++ solution (slower):
#include <iostream>

#include <cstdio>

#include <queue>

#include <string.h>



using namespace std;



#define MAXN 10001

#define INF 999999999





struct Edge

{

	int u, v, next, d, c;

}edge[MAXN * 4];



int head[MAXN], e;

int n, m;

int total;

int dis[MAXN];

int cost[MAXN];

bool vis[MAXN];



void add(int u, int v, int d, int c)

{

	edge[e].u = u;

	edge[e].v = v;

	edge[e].d = d;

	edge[e].c = c;

	edge[e].next = head[u];

	head[u] = e++;

}



void init()

{

	e = 0;

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

	memset(vis, false, sizeof(vis));

}



queue <int> q;



void solve()

{

	while (!q.empty())

	{

		q.pop();

	}

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

	{

		dis[i] = INF;

		cost[i] = INF;

	}

	

	int cur, next, ans = 0;

	q.push(1);

	vis[1] = true;

	dis[1] = 0;

	cost[1] = 0;

	

	while (!q.empty())

	{

		cur = q.front();

		q.pop();

		vis[cur] = false;

		for (int i = head[cur]; i != -1; i = edge[i].next)

		{

			int v = edge[i].v;

				if (dis[v] > dis[cur] + edge[i].d)

				{

					dis[v] = dis[cur] + edge[i].d;

					cost[v] = edge[i].c;

					if(!vis[v])

				     {

					vis[v] = true;

					q.push(v);

				    }

				}

				else if (dis[v] == dis[cur] + edge[i].d && cost[v] > edge[i].c)

	             {

			

					cost[v] = edge[i].c;

					if(!vis[v])

			     	{

					vis[v] = true;

					q.push(v);

				    }

				}

				

		}

		

		

	}

	

	int ss = 0;

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

	{

		ss += cost[i];

	}

	

	cout << ss << endl;

}



void input()

{

	int u, v, d, c;

	

	while (cin >> n >> m)

	{

		total = 0;

		if (!n && !m)

		{

			break;

		}

		init();

		

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

		{

			cin >> u >> v >> d >> c;

			add(u, v, d, c);

			add(v, u, d, c);

			total += 2 * c;

		}

		

		solve();

	}

}



int main()

{

	std::ios::sync_with_stdio(false);

	input();

	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: