// 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: 1.129323880795814x):
#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;

}