// 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: 1.2078519950600946x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

const int inf =0x3f3f3f3f;

const int MAXN=10005;

struct node

{

    int next, to, d, c;

} G[2000005];

int cur, n;

bool vis[MAXN];

int cost[MAXN];

int dist[MAXN];

int head[MAXN];

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

{

    G[cur].to=v;

    G[cur].c=c;

    G[cur].d=d;

    G[cur].next=head[u];

    head[u]=cur++;

}

queue<int>q;

void spfa(int s)

{



    while(!q.empty()) q.pop();

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

    fill(cost+1,cost+n+1,inf);

    fill(dist+1,dist+n+1,inf);

    q.push(s);

    dist[s]=0;

    cost[s]=0;

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=false;

        for(int i=head[u]; ~i; i=G[i].next)

        {

            int v=G[i].to;

            if(dist[v]>dist[u]+G[i].d||(dist[v]==dist[u]+G[i].d&&cost[v]>G[i].c))

            {

                dist[v]=dist[u]+G[i].d;

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

                if(!vis[v])

                {

                    vis[v]=true;

                    q.push(v);

                }

            }

        }

    }

}

int main()

{

    int m;

    int u, v, d, c;

    while(~scanf("%d %d", &n, &m))

    {

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

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

        cur=0;

        while(m--)

        {

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

            add(u, v, d, c);

            add(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;

}