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.io.PrintWriter;
import java.util.*;
public class test2 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		ArrayList<ArrayList<Integer>> adjmat = new ArrayList<ArrayList<Integer>>();
		TreeMap<Integer, TreeMap<Integer, Integer>>edj = new TreeMap();
		for(int i=0;i<n;i++) {
			adjmat.add(new ArrayList<Integer>());
			edj.put(i, new TreeMap<Integer, Integer>());
		}
		Pair[] edges = new Pair[n-1];
		for(int i=0;i<n-1;i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			edges[i] = new Pair(a-1, b-1);
			adjmat.get(a-1).add(b-1);
			adjmat.get(b-1).add(a-1);
		}
		if(n==2) {
			pw.println(0);
			
		}else {
		int c = 0;
		for(int i=0;i<adjmat.size();i++) {
			if(adjmat.get(i).size()==1) {
				edj.get(i).put(adjmat.get(i).get(0),c);
				edj.get(adjmat.get(i).get(0)).put(i, c);
				c++;
				
			}
		}
		for(int i=0;i<edges.length;i++) {
			if(edj.get(edges[i].a).get(edges[i].b)!=null)
			pw.println(edj.get(edges[i].a).get(edges[i].b));
			else pw.println(c++);
		}
		}
		pw.flush();
		pw.close();
	}
}
class Pair {
	int a,b;
	public Pair(int a,int b) {
		this.a = a;
		this.b = b;
	}
}