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

#include<queue>

#include<cstdlib>

#include<cstdio>

#include<cstring>



using namespace std;

const int maxn=4e4+100;

const int inf=0x3f3f3f3f;



struct Edge{

    int next;

    int to;

    int w;

    int  c;



}G[maxn];



int inq[maxn];

int dis[maxn];

int head[maxn];

int k=0;





void adde(int v,int to,int w,int c)//加?

{

    G[k].to=to;//第k条?指向的点

    G[k].w=w;//第k条?的??

    G[k].c=c;

    G[k].next=head[v];//与第k条??同起点的上一条?的?存位置

    head[v]=k++;

}



queue<int>Q;



void spfa(int s)

{

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

        memset(inq,0,sizeof(inq));

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

        dis[s] = 0;

        inq[s] = 1;

        Q.push(s);

        while(!Q.empty())

        {

                int u = Q.front();

                Q.pop();

                inq[u]=0;

                for(int i=head[u];i!=-1;i=G[i].next)//遍?以u?起点的所有?

                {

                        int v=G[i].to;

                        if(dis[v]>dis[u]+G[i].w)

                        {

                                dis[v]=dis[u]+G[i].w;

                                if(inq[v]==0)

                                {

                                    Q.push(v);

                                    inq[v]=1;

                                }

                        }

                }

                inq[u]=0;



        }

}



int main()

{

        int n,m;

        while(scanf("%d%d",&n,&m)!=EOF)

        {

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

                        return 0;

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



                k=0;

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

                {

                            int a,b,c,d;

                            scanf("%d%d%d%d",&a,&b,&c,&d);

                            adde(a,b,c,d);

                            adde(b,a,c,d);

                }

                spfa(1);

                int ans=0;

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

                {

                        int mi=inf;

                        for(int j=head[i];j!=-1;j=G[j].next)

                        {

                                if(dis[G[j].to]+G[j].w==dis[i]&&G[j].c<mi)

                                        mi=G[j].c;

                        }

                        ans+=mi;

                }

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

        }

        return 0;



}

// Faster C++ solution for the same problem (measured speedup: 1.6547919472257586x):
#include <bits/stdc++.h>

using namespace std;



#define db double

#define ll long long

#define RG register



inline int gi()

{

	RG int ret; RG bool flag; RG char ch;

	ret=0, flag=true, ch=getchar();

	while (ch < '0' || ch > '9')

		ch == '-' ? flag=false : 0, ch=getchar();

	while (ch >= '0' && ch <= '9')

		ret=(ret<<3)+(ret<<1)+ch-'0', ch=getchar();

	return flag ? ret : -ret;

}



const db pi = acos(-1.0);

const int N = 142857, inf = 1<<30;



int f[N],et[N],ed[N],nx[N],ew[N],dis[N],cst[N];

bool vis[N];

queue <int> q;

int num;



inline void add(int x,int y,int z,int w)

{

	et[++num]=y, nx[num]=f[x], ed[num]=z, ew[num]=w, f[x]=num;

}



int main()

{

	int n,m,i,x,y,z,w;

	while (true)

		{

			n=gi(), m=gi();

			if (n == 0)

				break;

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

				{

					x=gi(), y=gi(), z=gi(), w=gi();

					add(x,y,z,w), add(y,x,z,w);

				}

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

				cst[i]=dis[i]=inf;

			q.push(1), vis[1]=true;

			while (!q.empty())

				{

					x=q.front(), q.pop(), vis[x]=false;

					for (i=f[x]; i; i=nx[i])

						{

							y=et[i], z=ed[i], w=ew[i];

							if (dis[y] > dis[x]+z)

								{

									dis[y]=dis[x]+z;

									cst[y]=cst[x]+w;

									if (!vis[y])

										q.push(y), vis[y]=true;

								}

							if (dis[y] == dis[x]+z && cst[y] > w)

								cst[y]=w;

						}

				}

			x=0;

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

				x+=cst[i];

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

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

				f[i]=0;

			num=0;

		}

	return 0;

}