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

#include<algorithm>

#include<queue>

#include<iostream>

#include<cstring>

using namespace std;

const int maxn = 10050, maxe = 2005000, inf = 1e9;

struct node {

	int to;

	int w;

	int c;

	int next;

};

node edge[maxe];

int head[maxn];

int ip;

void init()

{

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

	ip = 0;

}

void addedge(int u, int v, int w, int c)

{

	edge[ip].to = v;

	edge[ip].c = c;

	edge[ip].w = w;

	edge[ip].next = head[u];

	head[u] = ip++;

}

int cost[maxn], dis[maxn], vis[maxn];

queue<int>q;

void spfa(int s, int n)

{

	while (!q.empty())

		q.pop();

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

		dis[i] = cost[i] = inf;

	memset(vis, 0, sizeof(vis));

	dis[s] = 0, cost[s] = 0, vis[s] = 1;

	q.push(s);

	while (q.size())

	{

		int u = q.front();

		q.pop();

		vis[u] = 0;

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

		{

			int to = edge[i].to, val = edge[i].w;

			if (dis[to] > dis[u] + val || (dis[to] == dis[u] + val&&cost[to] > edge[i].c))

			{

				dis[to] = dis[u] + val;

				cost[to] = edge[i].c;

				if (!vis[to])

				{

					q.push(to);

					vis[to] = 1;

				}

			}

		}

	}

}

int main()

{

	int n, m, u, v, w, c;

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

	{

		init();

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

		{

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

			addedge(u, v, w, c);

			addedge(v, u, w, c);

		}

		spfa(1, n);

		int ans = 0;

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

			ans += cost[i];

		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: