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

#include <queue>

#include <functional>

#include <cstring>

using namespace std;

#define MAX_V 10240

 

// 从?点from指向?点to的???cost的?

typedef struct edge

{

	int to, distance, cost;

	edge(){}

	edge(int to, int distance, int cost) : to(to), distance(distance), cost(cost){}

	bool operator > (const edge & b) const

	{

		return distance != b.distance  ? distance > b.distance : cost > b.cost;

	}

} P;

 

// first 最短路径，second?点?号

 

// ?

vector<edge> G[MAX_V];

 

// V是?点数

int V;

bool visited[MAX_V];

 

// 求解从?点s出?到所有点的最短花?

int dijkstra(int s)

{

	int result = 0;

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

	memset(visited, 0, V * sizeof(bool));

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

 

	while (!que.empty())

	{

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

		int v = p.to;

		if (visited[v]) continue;

		visited[v] = true;

		result += p.cost;

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

		{

			edge e = G[v][i];

			que.push(P(G[v][i].to, p.distance + G[v][i].distance, G[v][i].cost));

		}

	}

 

	return result;

}

 

///////////////////////////SubMain//////////////////////////////////

int main(int argc, char *argv[])

{

	int /*N = V,*/ M;

	while (cin >> V >> M && V)

	{

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

		{

			G[i].clear();

		}

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

		{

			int u, v, d, c;

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

			--u, --v;

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

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

		}

 

		cout << dijkstra(0) << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2810361090714686x):
#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

const int INF = 0x3f3f3f3f;

struct node

{

    int to, w, c, next;

} edge[50005];

int head[10005], ip, cost[10005], dis[10005], n, m;

bool vis[10005];

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

{

    edge[ip].to = v;

    edge[ip].w = d;

    edge[ip].c = c;

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

    head[u] = ip++;

}

void spfa(int s)

{

    queue<int> q;

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

        cost[i] = dis[i] = INF;

    memset(vis, 0, sizeof(vis));

    dis[s] = 0;

    cost[s] = 0;

    vis[s] = 1;

    q.push(s);

    while (!q.empty())

    {

        int u = q.front();

        q.pop();

        vis[u] = 0;

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

        {

            int to = edge[i].to;

            int w = edge[i].w;

            if(dis[to] > dis[u] + w||(dis[to] == dis[u]+w&&cost[to] > edge[i].c))

            {

                dis[to] = dis[u] + w;

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

                if(!vis[to])

                {

                    q.push(to);

                    vis[to] = 1;

                }

            }

        }

    }

}

int main()

{

    while (scanf("%d %d", &n, &m) != EOF)

    {

        if (n == 0 && m == 0)

            break;

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

        ip = 0;

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

        {

            int u, v, d, c;

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

            addedge(u, v, d, c);

            addedge(v, u, d, c);

        }

        spfa(1);

        int ans = 0;

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

            ans += cost[i];

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

    }

    return 0;

}