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

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <cstring>

#include <vector>

#include <cmath>

#include <queue>



using namespace std;

#define Fill(a, b) memset(a, b, sizeof(a))

#define Debug(x) cout<<#x<<"="<<(x)<<endl;

typedef pair<int, int> P;

const int inf = 0x1f1f1f1f;

const int max_n = 10010;

struct edge

{

    int from, to, cost, fee;

};

int n, ans;

//vector<int> g fee[max_n];

int d[max_n], pre[max_n];

vector<edge> G[max_n];

void dijkstra(int s);

vector<int> get_path(int t);

int main()

{

    int i = 0, j = 0, k = 0;

    int m, a, b, p, q;

    edge ei;

    while(scanf("%d %d", &n, &m)!=EOF)

    {

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

            break;

        ans = 0;

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

            G[i].clear();

        for(i = 0; i < m; i++)

        {

            scanf("%d %d %d %d", &a, &b, &p, &q);

            ei.from = a;

            ei.to = b;

            ei.cost = p;

            ei.fee = q;

            G[a].push_back(ei);

            ei.from = b;

            ei.to = a;

            G[b].push_back(ei);

        }

        dijkstra(1);



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

        /*

        for(i = 2; i <= n; i++)

            cout << pre[i] << " ";

        cout << endl;

        */

    }



    return 0;

}

void dijkstra(int s)

{

    int i, j, u, v;

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

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

    {

        d[i] = inf;

    }

    Fill(pre, -1);

    d[s] = 0;

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

    while(!que.empty())

    {

        P pa = que.top();

        que.pop();

        v = pa.second;

        if(d[v] < pa.first)

            continue;

        edge es;

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

        {

            es = G[v][i];

            if(d[es.to] > d[es.from]+es.cost)

            {

                if(d[es.to]!=inf)

                    ans -= pre[es.to];

                d[es.to] = d[es.from]+es.cost;

                ans += es.fee;

                pre[es.to] = es.fee;

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

            }

            else if((d[es.to]==d[es.from]+es.cost)&&(pre[es.to]>es.fee))

            {

                ans = ans - pre[es.to] + es.fee;

                pre[es.to] = es.fee;

            }

        }

    }

}

// Faster C++ solution for the same problem (measured speedup: 1.3207938023419485x):
#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;

}