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

#include<cstring>

#include<queue>

#include<vector>

using namespace std;

#define INF 1000000000

#define pb push_back

#define maxn 10001

struct Edge{

	int from, to, dist, cost;

};

struct HeapNode{

	int d, u, c;

	bool operator < (const HeapNode& a)const{

		if(d==a.d)	return c>a.c;

		return d>a.d;

	}

};

inline void IN(int& x){

	x=0;

	char c=getchar();

	while(c<48 || c>57)	c=getchar();

	while(c>=48 && c<=57){

		x=x*10+c-48;

		c=getchar();

	}

}

int n, m;

vector<Edge> edges;

vector<int> G[maxn];

vector<int> V[maxn];

bool done[maxn];

int d[maxn], p[maxn], c[maxn];

int ans;

void init(){

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

		G[i].clear();

		V[i].clear();

	}

	edges.clear();

}

void add(int from, int to, int dist, int cost){

	edges.pb((Edge){from, to, dist, cost});

	m = edges.size();

	G[from].pb(m-1);

	V[from].pb(cost);

}

void dijkstra(int s){

	ans=0;

	priority_queue<HeapNode> Q;

	for(int i=1; i<=n; i++)	d[i]=INF;

	for(int i=1; i<=n; i++)	c[i]=INF;

	d[s]=0;

	c[s]=0;

	memset(done, 0, sizeof(done));

	Q.push((HeapNode){0,s,0});

	while(!Q.empty()){

		HeapNode x = Q.top(); Q.pop();

		int u=x.u;

		if(done[u])	continue;

		done[u]=1;

		ans += c[u];

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

			Edge& e = edges[G[u][i]];

			if(d[e.to]>d[u]+e.dist || (d[e.to]==d[u]+e.dist && c[e.to]>V[u][i])){

				d[e.to] = d[u]+e.dist;

				p[e.to] = G[u][i];

				c[e.to] = V[u][i];

				Q.push((HeapNode){d[e.to],e.to,c[e.to]});

			}

		}

	}

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

}

int k;

int main(){

	while(1){

		IN(n); IN(k);

		if(!n && (!k))	break;

		int a, b, c, d;

		init();

		while(k--){

			IN(a); IN(b); IN(c); IN(d);

			add(a, b, c, d);

			add(b, a, c, d);

		}

		dijkstra(1);

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3420336836133042x):
#include <stdio.h>

#include <string.h>

#include <queue>

#include <iostream>

#include <algorithm>

using namespace std;

const int maxn=10050;

const int maxe=40414;

const int INF=0x3f3f3f3f;

struct note

{

    int v;

    int d;

    int c;

    int next;

};

note edge[maxe];

int head[maxn];

int cnt;

void init()

{

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

    cnt=0;

}

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

{

    edge[cnt].v=v;

    edge[cnt].c=c;

    edge[cnt].d=d;

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

    head[u]=cnt++;

}

int cost[maxn];

int dis[maxn];

bool vis[maxn];

queue<int>q;

void spfa(int n)

{

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

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

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

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

    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)

        {

            int v=edge[i].v;

            int val=edge[i].d;

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

            {

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

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

                if(!vis[v])

                {

                    q.push(v);

                    vis[v]=true;

                }

            }

        }

    }

}

int main()

{

    int n,m;

    int u,v,d,c;

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

    {

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

        init();

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

        {

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

            addedge(u,v,c,d);

            addedge(v,u,c,d);

        }

        spfa(n);

        int ans=0;

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

            ans+=cost[i];

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

    }

    return 0;

}