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>
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        int n=ni();
        //ArrayList<Integer>[]A=new ArrayList[n+1];
        ArrayList<Integer>[]B=new ArrayList[n+1];
        for (int x=1;x<=n;x++) B[x]=new ArrayList();
        for (int x=1;x<n;x++) {
            int u=ni(),v=ni();
            B[u].add(x);
            B[v].add(x);
        }
        
        int[]C=new int[n];
        Arrays.fill(C,-1);
        boolean f=false;
        for (int x=1;x<=n;x++) {
            if (B[x].size()<3) continue;
            for (int y=0;y<3;y++) C[B[x].get(y)]=y;
            f=true;
            break;
        }
        
        int p=0;
        if (f) p=3;
        
        for (int x=1;x<n;x++) {
            if (C[x]==-1) { out.println(p); p++; }
            else out.println(C[x]);
        }
        out.flush();
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}