// Original C++ solution (slower):
#include<iostream>

#include<queue>

#include<functional>

#include<cstring>



using namespace std;



#define maxn 10240 

#define INF 0x3f3f3f3f;



typedef struct edge{

	int to,dis,cos;

	edge(){}

	edge(int to,int dis,int cos):to(to),dis(dis),cos(cos){}

};

typedef pair<int ,int> P;



vector<edge>G[maxn];



int d[maxn];

int v;//定点



void dijkstra(int s){

	priority_queue<P,vector<P>,greater<P> > que;//优先输出小数据；

	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(d[v]<p.first)continue;

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

		{

			edge e=G[v][i];

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

			{

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

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

			}

		}

	}	

} 

int main()

{

	int m;

	while(cin>>v>>m)

	{

		if(v==0&&m==0)break;

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

		{

			G[i].clear();

		}

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

		{

			int u,v,d,c;

			cin>>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<v;i++)

		{

			int min=INF;

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

			{

				if(d[G[i][j].to]+G[i][j].dis==d[i]&&G[i][j].cos<min)

				{

					min=G[i][j].cos;

				}

				

			} 

			ans+=min;

		}cout<<ans<<endl;

	}

	return 0;

}


// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#define _CRT_SBCURE_NO_DEPRECATE

#include <set>

#include <map>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <functional>

//#define UP(i,x,y) for(int i=x;i<=y;i++)

//#define DOWN(i,x,y) for(int i=x;i>=y;i--)

#define sdddd(x,y,z,k) scanf("%d%d%d%d", &x, &y, &z, &k)

#define sddd(x,y,z) scanf("%d%d%d", &x, &y, &z)

#define sdd(x,y) scanf("%d%d", &x, &y)

#define sd(x) scanf("%d", &x)

#define mp make_pair

#define pb push_back

#define lson k<<1

#define rson k<<1|1

#define mid (l+r)/2

#define ms(x, y) memset(x, y, sizeof x)

using namespace std;

typedef long long ll;

//typedef unsigned long long ull;

#define MOD 142857

const int maxn = 10100;

const int INF = 0x3f3f3f3f;

const ll LINF = 0x3f3f3f3f3f3f3f3f;

int n, m, s, t;

int len;

struct road

{

	int u, v, len, cost;

	int next;

};

struct node

{

    int v;

    int len;

    int cost;

    node(int v, int l, int c):v(v),len(l),cost(c){}

    friend bool operator <(node a, node b)

    {

        return a.len>b.len;

    }

};

road G[maxn*5];

int h[maxn];

int mx = -INF;

int d[maxn];

int c[maxn];

int path[maxn];

void dij()

{

    fill(d, d+n+1, INF);

    fill(path, path+n+1, -1);

    fill(c, c+n+1, 0);

    d[1] = 0;

    priority_queue<node> pque;

    pque.push(node(1, 0, 0));

    while(pque.size())

    {

        node p = pque.top(); pque.pop();

        int v = p.v;

        if(d[v] < p.len) continue;

        for(int i = h[v]; i != -1; i = G[i].next)

        {

            road e = G[i];

            if(d[e.v] > d[e.u] + e.len)

            {

                d[e.v] = d[e.u] + e.len;

                c[e.v] = c[e.u] + e.cost;

                path[e.v] = i;

                pque.push(node(e.v, d[e.v], c[e.v]));

            }

            else if(d[e.v] == d[e.u] + e.len)

            {

                if(path[e.v] != -1 && G[path[e.v]].cost > e.cost)

                {

                    path[e.v] = i;

                    pque.push(node(e.v, d[e.v], c[e.v]));

                }

            }

        }

    }

}

int main()

{

	//freopen("out.txt", "w", stdout);

    while(~sdd(n, m) && (n||m))

    {

        int ta, tb, tc, td;

        int len = 0;

        fill(h, h+n+1, -1);

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

        {

            sdddd(ta, tb, tc, td);

            G[len].u = ta;

            G[len].v = tb;

            G[len].len = tc;

            G[len].cost = td;

            G[len].next = h[ta];

            h[ta] = len++;



            G[len].u = tb;

            G[len].v = ta;

            G[len].len = tc;

            G[len].cost = td;

            G[len].next = h[tb];

            h[tb] = len++;

        }

        dij();

        int ans = 0;

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

        {

            if(path[i]!=-1)

            {

                ans += G[path[i]].cost;

            }

        }

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

    }



	return 0;

}




// Fastest known solution for this problem: