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

#include <queue>

#include <functional>

#include <cstring>

using namespace std;



struct edge{

    int v, d, c;

    edge(){}

    edge(int v, int d, int c) : v(v),d(d),c(c){}

};



int N, M;

const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;//距?， ?点

int d[10665];



vector<edge> G[10665];



void dijkstra(int s){

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

    memset(d, INF, sizeof(d));

    d[s] = 0;

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

    while(!que.empty())

    {

        P p = que.top();

        que.pop();

        int q = p.second;

        if(d[q] < p.first) continue;

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

        {

            edge e = G[q][i];

            if(d[e.v] > d[q] + e.d)

            {

                d[e.v] = d[q] + e.d;

                que.push(P(d[e.v], e.v));

            }

        }

    }

}





int main()

{

    while(cin >> N >> M)

    {

        if(!N) break;

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

		{

			G[i].clear();

		}

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

        {

            int u,t,c,d;

            cin >> u >> t >> d >> c;

            u--;t--;

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

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

        }

        dijkstra(0);

        int ans=0;

        for(int k = 1; k < N; k++)

        {

            int mincost = INF;

            for(int j = 0; j < G[k].size(); j++)

            {

                if(d[G[k][j].v]+G[k][j].d == d[k]

                 && G[k][j].c < mincost)

                    mincost = G[k][j].c;

            }

            ans += mincost;

        }

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1833502376456408x):
#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<cmath>

#include<algorithm>

#include<cstring>

#include<queue>

#define IL inline

#define RG register

#define inf 0x3f3f3f3f

#define LL long long

#define N 10005

#define M 20005

using namespace std;

IL int gi()

{

	RG int res=0,s=1;RG char ch;

	for(ch=getchar(); (ch<'0'||ch>'9')&&ch!='-' ;ch=getchar());

	if(ch=='-') s=-1,ch=getchar();

	for(;ch>='0'&&ch<='9';ch=getchar()) res=10*res+ch-48;

	return res*s;

}

int n,m,head[N],edge[2*M][5],num=-1;

int dis[N],flag[N],ans=0,a[N];

struct node

{

	int d,v;

	node(RG int dd=0,RG int vv=0){d=dd,v=vv;}

	bool operator < (const node &o) const{

		return v>o.v;

	}

};

priority_queue<node>heap;

IL void adde(RG int u,RG int v,RG int d,RG int c)

{

	edge[++num][0]=v;edge[num][1]=head[u];head[u]=num;edge[num][2]=d,edge[num][3]=c,edge[num][4]=0;

	edge[++num][0]=u;edge[num][1]=head[v];head[v]=num;edge[num][2]=d,edge[num][3]=c,edge[num][4]=0;

}

IL void init(){

	memset(head,-1,sizeof(head)); num=-1; ans=0;

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

	memset(flag,0,sizeof(flag));

}

IL int cmp(RG int x,RG int y){return dis[x]<dis[y];}

IL void work1()

{

	RG int p;

	dis[1]=0; heap.push(node(1,0));

	while(!heap.empty())

		{

			p=heap.top().d; heap.pop();

			while(flag[p]&&!heap.empty()) p=heap.top().d,heap.pop();

			if(flag[p]) break; flag[p]=1;

			for(RG int i=head[p];i!=-1;i=edge[i][1]){

				if(dis[edge[i][0]]>dis[p]+edge[i][2]){

					dis[edge[i][0]]=dis[p]+edge[i][2];

					heap.push(node(edge[i][0],dis[edge[i][0]]));

				}

			}

		}

	RG int mn;

	for(RG int i=1;i<=n;++i) a[i]=i;

	sort(a+1,a+n+1,cmp);

	for(RG int i=2;i<=n;++i)

		{

			mn=inf;

			for(RG int j=head[i];j!=-1;j=edge[j][1])

				{

					if(dis[i]==dis[edge[j][0]]+edge[j][2])

						mn=min(mn,edge[j][3]);

				}

			ans+=mn;

		}

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

}

int main()

{

	//freopen("1.in","r",stdin);

	//freopen("1.out","w",stdout);

	RG int u,v,d,c;

	while(true)

		{

			n=gi(),m=gi(); if(!n&&!m) break; init();

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

				u=gi(),v=gi(),d=gi(),c=gi(),adde(u,v,d,c);

			work1();

		}

	//fclose(stdin);fclose(stdout);

	return 0;

}