// 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: 1.1126285952900337x):
#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>



using namespace std;



const int inf = 0x3f3f3f3f;

const int N = 10414;

const int M = 40414;



struct Edge{

	int u, v, d, c;

	int next;

}edge[M];



int cnt, head[N];

int op, tp, link[N];

int vis[N], dis[N], cost[N];

queue<int>q;



void add_edge(int u, int v, int d, int c);

void spfa(int n);



int main(){

	int n, m, i, u, v, d, c;

	while(scanf("%d %d", &n, &m) && (n || m)){

		cnt = 0;

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

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

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

			add_edge(u, v, d, c);

			add_edge(v, u, d, c);

		}

		spfa(n);

	}

	return 0;

}

void add_edge(int u, int v, int d, int c){

	edge[cnt].u = u;

	edge[cnt].v = v;

	edge[cnt].d = d;

	edge[cnt].c = c;

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

	head[u] = cnt++;

}

void spfa(int n)

{

    int v, d, c;

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

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

    memset(dis, inf, sizeof(dis));

    memset(cost, inf, sizeof(cost));

    dis[1]=0;

    cost[1]=0;

    vis[1]=true;

    q.push(1);

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=false;

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

        {

            v = edge[i].v;

            d = edge[i].d;

            c = edge[i].c;

            if(dis[v] > dis[u] + d||(dis[v] == dis[u] + d && cost[v] > c))//有多条最短路时，取花费最小的

            {

                dis[v] = dis[u] +d;

                cost[v] = c;

                if(!vis[v])

                {

                    q.push(v);

                    vis[v]=true;

                }

            }

        }

    }

    int sum = 0;

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

    	sum += cost[i];

    }

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

    return;

}