// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



typedef long long ll;

const int N = 10005;

const int M = 1000000007;



int n, m;

typedef int wtype;

struct Edge { int v; wtype d, c; };

vector<Edge> e[N];

wtype dist[N], cost[N];

int pre[N];

bool vis[N];

int SPFA(int src) {

  memset(dist, 0x3f, sizeof(dist)); dist[src] = 0;

  memset(cost, 0x3f, sizeof(cost)); cost[src] = 0;

  memset(vis, 0, sizeof(vis)); vis[src] = true;

  stack<int> que; que.push(src);

  while (!que.empty()) {

    int u = que.top(); que.pop();

    vis[u] = false;

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

      int v = e[u][i].v; wtype d = e[u][i].d, c = e[u][i].c;

      if (dist[v] > dist[u] + d || (dist[v] == dist[u] + d && cost[v] != 0x3f3f3f3f && cost[v] > c)) {

        dist[v] = dist[u] + d; cost[v] = c;

        if (!vis[v]) { vis[v] = true; que.push(v); }

      }

    }

  }

  int ret = 0;

  for (int i = 1; i <= n; i++) { ret += cost[i]; }

  return ret;

}



int main() {

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

    for (int i = 1; i <= n; i++) { e[i].clear(); }

    for (int i = 0, u, v, d, c; i < m; i++) {

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

      e[u].push_back((Edge) {v, d, c});

      e[v].push_back((Edge) {u, d, c});

    }

    printf("%d\n", SPFA(1));

  }

}

// Faster C++ solution for the same problem (measured speedup: 1.2889570996841966x):
#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;

}