// 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: 1.344788807594167x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

const int inf =0x3f3f3f3f;

const int MAXN=10005;

struct node

{

    int next, to, d, c;

} G[2000005];

int cur, n;

bool vis[MAXN];

int cost[MAXN];

int dist[MAXN];

int head[MAXN];

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

{

    G[cur].to=v;

    G[cur].c=c;

    G[cur].d=d;

    G[cur].next=head[u];

    head[u]=cur++;

}

queue<int>q;

void spfa(int s)

{



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

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

    fill(cost+1,cost+n+1,inf);

    fill(dist+1,dist+n+1,inf);

    q.push(s);

    dist[s]=0;

    cost[s]=0;

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=false;

        for(int i=head[u]; ~i; i=G[i].next)

        {

            int v=G[i].to;

            if(dist[v]>dist[u]+G[i].d||(dist[v]==dist[u]+G[i].d&&cost[v]>G[i].c))

            {

                dist[v]=dist[u]+G[i].d;

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

                if(!vis[v])

                {

                    vis[v]=true;

                    q.push(v);

                }

            }

        }

    }

}

int main()

{

    int m;

    int u, v, d, c;

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

    {

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

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

        cur=0;

        while(m--)

        {

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

            add(u, v, d, c);

            add(v, u, d, c);

        }

        spfa(1);

        int ans=0;

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

            ans+=cost[i];

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

    }

    return 0;

}