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



using namespace std;

typedef pair<int, int> P;

typedef long long ll;



const int maxv = 10000 + 10;

struct edge

{

    int to, d, c;

    edge() {}

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

};

vector<edge> G[maxv];

int d[maxv];



int N, M;



void dijkstra(int s)

{

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

    memset(d, 0x3f, sizeof(d));

    d[s] = 0;

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

    while(!que.empty())

    {

        P p = que.top(); que.pop();

        int v = p.second;

        if(p.first > d[v]) continue;

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

        {

            edge e = G[v][i];

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

            {

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

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

            }

        }

    }

}



int main()

{

    while(scanf("%d%d", &N, &M)==2 && N)

    {

        for(int i=0; i<N; i++)  G[i].clear();

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

        {

            int u, v, d, c;

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

            u--; v--;

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

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

        }

        dijkstra(0);

        int ans = 0;

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

        {

            int mincost = 0x3f3f3f3f;

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

            {

                edge e = G[i][j];

                if(d[e.to]+e.d==d[i] && e.c<mincost)

                    mincost = e.c;

            }

            ans += mincost;

        }

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

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2486347111696838x):
#include <cstdio>

#include <cstdlib>

#include <iostream>

#include <cstring>

#include <cmath>

#include <ctime>

#include <algorithm>

#include <vector>

#include <set>

#include <queue>

#include <map>

#include <stack>

#include <bitset>

using namespace std;

typedef int gitype;

gitype gi () {

    gitype ret=0; char ch=getchar();

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

    char c=ch=='-'?getchar():ch;

    while(c>='0' && c<='9') ret=ret*10+c-'0',c=getchar();

    return ch=='-'?-ret:ret;

}

const int N = 10010;

const int M = 100010;

int n,m,st[N],vis[N],dis[N],pt[M],nxt[M],w[M],c[M],cost[N],ent=0;

void link (int u,int v,int _w,int _c) {

    pt[++ent]=v; nxt[ent]=st[u]; st[u]=ent; w[ent]=_w; c[ent]=_c;

    pt[++ent]=u; nxt[ent]=st[v]; st[v]=ent; w[ent]=_w; c[ent]=_c;

}

queue<int> q;

int main () {

    while(scanf("%d%d",&n,&m)==2 && (n || m)) {

        memset(st,0,sizeof(st)); ent=0;

        int u,v,_w,_c;

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

            u=gi(); v=gi(); _w=gi(); _c=gi(); link(u,v,_w,_c);

        }

        for(int i=1;i<=n;i+=1) dis[i]=1e9 + 5;

        dis[1]=0; q.push(1);

        while(!q.empty()) {

            int r=q.front(); q.pop(); vis[r]=false;

            for(int i=st[r];i;i=nxt[i]) {

                if(dis[pt[i]] > dis[r]+w[i]) {

                    dis[pt[i]] = dis[r]+w[i]; cost[pt[i]] = c[i];

                    if(!vis[pt[i]]) {

                        vis[pt[i]] = true; q.push(pt[i]);

                    }

                } else if(dis[pt[i]] == dis[r]+w[i]) {

                    cost[pt[i]] = min(cost[pt[i]], c[i]);

                }

            }

        }

        int ans = 0;

        for(int i=2;i<=n;i+=1) ans += cost[i];

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

    }

    return 0;

}