// 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: 1.3290108679179304x):
#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;

}