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

#include<queue>

#include<cstdlib>

#include<cstdio>

#include<cstring>



using namespace std;

const int maxn=4e4+100;

const int inf=0x3f3f3f3f;



struct Edge{

    int next;

    int to;

    int w;

    int  c;



}G[maxn];



int inq[maxn];

int dis[maxn];

int head[maxn];

int k=0;





void adde(int v,int to,int w,int c)//加?

{

    G[k].to=to;//第k条?指向的点

    G[k].w=w;//第k条?的??

    G[k].c=c;

    G[k].next=head[v];//与第k条??同起点的上一条?的?存位置

    head[v]=k++;

}



queue<int>Q;



void spfa(int s)

{

        while(!Q.empty())Q.pop();

        memset(inq,0,sizeof(inq));

        memset(dis,inf,sizeof(dis));

        dis[s] = 0;

        inq[s] = 1;

        Q.push(s);

        while(!Q.empty())

        {

                int u = Q.front();

                Q.pop();

                inq[u]=0;

                for(int i=head[u];i!=-1;i=G[i].next)//遍?以u?起点的所有?

                {

                        int v=G[i].to;

                        if(dis[v]>dis[u]+G[i].w)

                        {

                                dis[v]=dis[u]+G[i].w;

                                if(inq[v]==0)

                                {

                                    Q.push(v);

                                    inq[v]=1;

                                }

                        }

                }

                inq[u]=0;



        }

}



int main()

{

        int n,m;

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

        {

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

                        return 0;

                memset(head,-1,sizeof(head));



                k=0;

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

                {

                            int a,b,c,d;

                            scanf("%d%d%d%d",&a,&b,&c,&d);

                            adde(a,b,c,d);

                            adde(b,a,c,d);

                }

                spfa(1);

                int ans=0;

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

                {

                        int mi=inf;

                        for(int j=head[i];j!=-1;j=G[j].next)

                        {

                                if(dis[G[j].to]+G[j].w==dis[i]&&G[j].c<mi)

                                        mi=G[j].c;

                        }

                        ans+=mi;

                }

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

        }

        return 0;



}

// Faster C++ solution for the same problem (measured speedup: 1.6258018033277457x):
//#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;

}