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

#include <queue>

#include <functional>

#include <cstring>

using namespace std;



struct edge{

    int v, d, c;

    edge(){}

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

};



int N, M;

const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;//距?， ?点

int d[10665];



vector<edge> G[10665];



void dijkstra(int s){

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

    memset(d, INF, sizeof(d));

    d[s] = 0;

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

    while(!que.empty())

    {

        P p = que.top();

        que.pop();

        int q = p.second;

        if(d[q] < p.first) continue;

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

        {

            edge e = G[q][i];

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

            {

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

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

            }

        }

    }

}





int main()

{

    while(cin >> N >> M)

    {

        if(!N) break;

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

		{

			G[i].clear();

		}

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

        {

            int u,t,c,d;

            cin >> u >> t >> d >> c;

            u--;t--;

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

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

        }

        dijkstra(0);

        int ans=0;

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

        {

            int mincost = INF;

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

            {

                if(d[G[k][j].v]+G[k][j].d == d[k]

                 && G[k][j].c < mincost)

                    mincost = G[k][j].c;

            }

            ans += mincost;

        }

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3664722988997036x):
#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;

}