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

#include <queue>

#include <functional>

#include <cstring>

using namespace std;

#define MAX_V 10240

 

// 从?点from指向?点to的???cost的?

typedef struct edge

{

	int to, distance, cost;

	edge(){}

	edge(int to, int distance, int cost) : to(to), distance(distance), cost(cost){}

	bool operator > (const edge & b) const

	{

		return distance != b.distance  ? distance > b.distance : cost > b.cost;

	}

} P;

 

// first 最短路径，second?点?号

 

// ?

vector<edge> G[MAX_V];

 

// V是?点数

int V;

bool visited[MAX_V];

 

// 求解从?点s出?到所有点的最短花?

int dijkstra(int s)

{

	int result = 0;

	priority_queue<P, vector<P>, greater<P> > que;

	memset(visited, 0, V * sizeof(bool));

	que.push(P(0, 0, 0));

 

	while (!que.empty())

	{

		P p = que.top(); que.pop();

		int v = p.to;

		if (visited[v]) continue;

		visited[v] = true;

		result += p.cost;

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

		{

			edge e = G[v][i];

			que.push(P(G[v][i].to, p.distance + G[v][i].distance, G[v][i].cost));

		}

	}

 

	return result;

}

 

///////////////////////////SubMain//////////////////////////////////

int main(int argc, char *argv[])

{

	int /*N = V,*/ M;

	while (cin >> V >> M && V)

	{

		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));

		}

 

		cout << dijkstra(0) << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2252167152378055x):
#include <cstdio>

#include <cstdlib>

#include <iostream>

#include <cstring>

#include <cmath>

#include <ctime>

#include <algorithm>

#include <vector>

#include <set>

#include <queue>

#include <map>

#include <stack>

#include <bitset>

using namespace std;

typedef int gitype;

gitype gi () {

    gitype ret=0; char ch=getchar();

    while((ch<'0' || ch>'9') && ch!='-') ch=getchar();

    char c=ch=='-'?getchar():ch;

    while(c>='0' && c<='9') ret=ret*10+c-'0',c=getchar();

    return ch=='-'?-ret:ret;

}

const int N = 10010;

const int M = 100010;

int n,m,st[N],vis[N],dis[N],pt[M],nxt[M],w[M],c[M],cost[N],ent=0;

void link (int u,int v,int _w,int _c) {

    pt[++ent]=v; nxt[ent]=st[u]; st[u]=ent; w[ent]=_w; c[ent]=_c;

    pt[++ent]=u; nxt[ent]=st[v]; st[v]=ent; w[ent]=_w; c[ent]=_c;

}

queue<int> q;

int main () {

    while(scanf("%d%d",&n,&m)==2 && (n || m)) {

        memset(st,0,sizeof(st)); ent=0;

        int u,v,_w,_c;

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

            u=gi(); v=gi(); _w=gi(); _c=gi(); link(u,v,_w,_c);

        }

        for(int i=1;i<=n;i+=1) dis[i]=1e9 + 5;

        dis[1]=0; q.push(1);

        while(!q.empty()) {

            int r=q.front(); q.pop(); vis[r]=false;

            for(int i=st[r];i;i=nxt[i]) {

                if(dis[pt[i]] > dis[r]+w[i]) {

                    dis[pt[i]] = dis[r]+w[i]; cost[pt[i]] = c[i];

                    if(!vis[pt[i]]) {

                        vis[pt[i]] = true; q.push(pt[i]);

                    }

                } else if(dis[pt[i]] == dis[r]+w[i]) {

                    cost[pt[i]] = min(cost[pt[i]], c[i]);

                }

            }

        }

        int ans = 0;

        for(int i=2;i<=n;i+=1) ans += cost[i];

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

    }

    return 0;

}