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

#include <cstring>

#include <queue>

#define inf 0x3f3f3f3f

#define maxm 10005

#define maxn 40005



using namespace std;



int m,n;

long long edgenum,head[maxm],dis[maxm],num[maxm],ans;

bool vis[maxm];



struct node

{

    int from,to,len,cost,next;

}edge[maxn];



void init()

{

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

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

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

    memset(num,inf,sizeof(num));

    edgenum=0;

}



void add(int u,int v,int len,int cost)

{

    node tp={u,v,len,cost,head[u]};

    edge[edgenum]=tp;

    head[u]=edgenum++;

}



long long SPFA(int st)

{

     queue<int> q;

     q.push(st);

     vis[st]=1;

     dis[st]=0;

     num[st]=0;

     ans=0;

     while(!q.empty())

     {

          int tmp=q.front();

          q.pop();

          vis[tmp]=0;

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

          {

                if((dis[edge[i].to]>dis[tmp]+edge[i].len)||((dis[edge[i].to]==dis[tmp]+edge[i].len)&&(num[edge[i].to]>edge[i].cost)))

                {

                    dis[edge[i].to]=dis[tmp]+edge[i].len;

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

                    if(!vis[edge[i].to])

                    {

                        q.push(edge[i].to);

                        vis[edge[i].to]=1;

                    }

                }

          }

     }

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

        ans+=num[i];

     return ans;

}



int main()

{

    int u,v,len,cost;

    while(cin>>m>>n&&m)

    {

        init();

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

        {

            cin>>u>>v>>len>>cost;

            add(u,v,len,cost);

            add(v,u,len,cost);

        }

        cout<<SPFA(1)<<endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1715862932448644x):
#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;

}