using namespace std;
#include<iostream>
#include<fstream>
#include<vector>

#define nmax 200010
#define pb push_back
#define oo 0x3f3f3f3f
struct edge{
    long long x,y;

    int ind;
    int c1[20];
    int c2[20];
    int p[40];
    int isneg;
};
edge a[nmax];
int T[nmax],r[nmax];
vector<int> ans;
int N,M;
ofstream fout("lazy.out");

void init()
{
    int i;
    for(i=1;i<=N;i++)
    {
        T[i]=i;
        r[i]=1;
    }
}

int find(int x)
{
    int R,y;
    for(R=x;R!=T[R];R=T[R]);
    for(;x!=T[x];)
    {
        y=T[x];
        T[x]=R;
        x=y;
    }
    return R;
}

void unite(int x,int y)
{
    if(r[x]>r[y])
    {
        T[y]=x;

    }
    else
    {
        T[x]=y;
    }
    if(r[x]==r[y]) r[y]++;
}

void solve()
{
    int i;
    for(i=1;i<=M;i++)
    {
        if(find(a[i].x)!=find(a[i].y))
        {
            unite(find(a[i].x),find(a[i].y));
            ans.pb(a[i].ind);
        }
    }
    vector<int>::iterator it;
    for(it=ans.begin();it<ans.end();it++)
    {
        fout<<*it<<"\n";
    }
}

bool cmp(edge i,edge j)
{
     if(i.isneg==1 && i.isneg=0)
        return 1;
     if(
     return 0;
}
void mul(int A[], int B[])
{
      int i, j, t, C[NR_CIFRE];
      memset(C, 0, sizeof(C));
      for (i = 1; i <= A[0]; i++)
      {
              for (t=0, j=1; j <= B[0] || t; j++, t/=10)
                      C[i+j-1]=(t+=C[i+j-1]+A[i]*B[j])%10;
              if (i + j - 2 > C[0]) C[0] = i + j - 2;
      }
      memcpy(A, C, sizeof(C));
}
void cit()
{
    ifstream fin("lazy.in");
    int i,j,x,y;
    long long c1,c2;
    fin>>N>>M;
    for(i=1;i<=M;i++)
    {
        fin>>a[i].x>>a[i].y>>c1>>c2;
        //a[i].c=c1;

        while(c1)
        {
          a[i].c1[++a[i].c1[0]]=c1%10;
          c1/=10;
        }

       // a[i].p=c1*c2;
       if(c2<0)
         a[i].isneg=1,c2=-c2;
        while(c2)
        {
          a[i].c2[++a[i].c2[0]]=c2%10;
          c2/=10;
        }
        mul(a[i].c2,a[i].c1);
        a[i].ind=i;
    }


    fin.close();

}


int main()
{
    cit();
/*    unsigned long long prod=1;
    int i;
    for(i=1;i<=62;i++)
     prod*=2;
     long long p2=(1<<17);
     double f;
     f=(double)(prod/p2);
    //fout<<prod<<" " <<p2<<" "<<f<<"\n";
*/
    sort(a+1,a+M+1,cmp);
    init();
    solve();
    fout.close();
    return 0;
}