Problem: You are given a tree consisting of n nodes. You want to write some labels on the tree's edges such that the following conditions hold:

  * Every label is an integer between 0 and n-2 inclusive. 
  * All the written labels are distinct. 
  * The largest value among MEX(u,v) over all pairs of nodes (u,v) is as small as possible. 



Here, MEX(u,v) denotes the smallest non-negative integer that isn't written on any edge on the unique simple path from node u to node v.

Input

The first line contains the integer n (2 ≤ n ≤ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines contains two space-separated integers u and v (1 ≤ u,v ≤ n) that mean there's an edge between nodes u and v. It's guaranteed that the given graph is a tree.

Output

Output n-1 integers. The i^{th} of them will be the number written on the i^{th} edge (in the input order).

Examples

Input


3
1 2
1 3


Output


0
1


Input


6
1 2
1 3
2 4
2 5
5 6


Output


0
3
2
4
1

Note

The tree from the second sample:

<image>

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Main implements Runnable{
    FastScanner sc;
    PrintWriter pw;
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args)
    {
        new Thread(null,new Main(),"codeforces",1<<25).start();
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
       int t=1;
       while(t-->0)
       solve();
        pw.flush();
        pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////
    public void solve()
    {
       int n=sc.ni();
       int[] deg=new int[n];
       int[] arr=new int[n];
       int[] brr=new int[n];
       for(int i=0;i<n-1;i++)
       {
           int x=sc.ni()-1;
           int y=sc.ni()-1;
           deg[x]++;
           deg[y]++;
           arr[i]=x;
           brr[i]=y;
       }
       int[] col=new int[n-1];
       if(n==2)
       {
           pw.println("0");
           return;
       }
       Arrays.fill(col,-1);
       int tmp=0;
       for(int i=0;i<n-1;i++)
       {
           if(deg[arr[i]]==1||deg[brr[i]]==1)
           col[i]=tmp++;
       }
       for(int i=0;i<n-1;i++)
       {if(col[i]==-1)
       col[i]=tmp++;
           pw.println(col[i]);
       }
    }
}