// Original C++ solution (slower):
#include<stdio.h>

#include<string.h>  

#include<math.h>  

  

//#include<map>   

//#include<set>

#include<deque>  

#include<queue>  

#include<stack>  

#include<bitset> 

#include<string>  

#include<iostream>  

#include<algorithm>  

using namespace std;  

  

#define ll long long  

#define INF 0x3f3f3f3f  

#define mod 1000000

#define clean(a,b) memset(a,b,sizeof(a))// 水印 



struct node{

	int to,next,dis,cost;

}edge[40100];

int head[10100];

int dis[10100],cost[10100];

bool vst[10100];

int n,m,k;



void spfa(int x)

{

	clean(dis,INF);

	clean(cost,INF);

	clean(vst,0);

	queue<int> q;

	dis[x]=0;

	cost[x]=0;

	q.push(x);

	while(!q.empty())

	{

		x=q.front();q.pop();

		vst[x]=0;

		for(int i=head[x];i!=-1;i=edge[i].next)

		{

			node b=edge[i];

			if(dis[b.to]>dis[x]+b.dis)

			{

				if(!vst[b.to])

				{

					vst[b.to]=1;

					q.push(b.to);

				}

				dis[b.to]=dis[x]+b.dis;

				cost[b.to]=b.cost;

			}

			else if(dis[b.to]==dis[x]+b.dis)

			{

				if(cost[b.to]>b.cost)

					cost[b.to]=b.cost;

			}

		}

	}

}



void add(int a,int b,int dis,int cost)

{

	edge[k].to=b;

	edge[k].dis=dis;

	edge[k].cost=cost;

	edge[k].next=head[a];

	head[a]=k;

	k++;

}



int main()

{

	while(cin>>n>>m&&n!=0||m!=0)

	{

		clean(head,-1);

		k=0;

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

		{

			int a,b,d,c;

			cin>>a>>b>>d>>c;

			add(a,b,d,c);

			add(b,a,d,c);

		}

		spfa(1);

		ll res=0;

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

			res=res+cost[i];

		cout<<res<<endl;

	}

}


// 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: