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

#include <iostream>

#include <cstring>

#include <string>

#include <algorithm>

#include <vector>

#include <queue>

const int maxn = 10007;

const int inf =1e9;

using namespace std;

struct node{

	int to,d,c;

	node(){}

	node (int to,int d,int c):to(to),d(d),c(c){}

};

vector<node>mp[maxn];

int dis[maxn],cost[maxn],n,m,ans = 0;

void init(){

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

	{

		mp[i].clear();

		dis[i] = inf;

		cost[i]  = 0;

	}

}

void add(int u,int v,int d,int c)

{

	mp[u].push_back(node(v,d,c));

}

int main(){

	while(~scanf("%d%d",&n,&m))

	{

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

		init();

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

		{

			int u,v,d,c;

			scanf("%d%d%d%d",&u,&v,&d,&c);

			add(u,v,d,c);

			add(v,u,d,c);

		}

		priority_queue<pair<int,int> >q;

		dis[1]=0;

		cost[1]=0;

		

		q.push (make_pair(-dis[1],1)) ;

		while(!q.empty())

		{

			int now = q.top().second;

			q.pop();

			for(int i=0;i<mp[now].size();i++)

			{

				int v=mp[now][i].to;

				int d=mp[now][i].d;

				int c=mp[now][i].c;

				if(dis[v]>dis[now]+d)

				{

					dis[v] = dis[now]+d;

					//ans -=cost[v];

					//cost[v] = cost[now]+c;

					//ans +=cost[v];

					q.push (make_pair(-dis[v],v));

				}

			}

		}

		ans = 0;

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

		{

			int minc = inf;

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

			{

				node tmp = mp[i][j];

				if(tmp.d+dis[tmp.to]==dis[i]&&minc>tmp.c)

				{

					minc = mp[i][j].c;

				}

			}

			ans += minc;

		}

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

	}

	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: