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:

//package codeforce;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
 
 
public class submit {
    public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		ArrayList<Integer> arl[] = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			arl[i] = new ArrayList<>();
		}
		for (int i = 0; i < n - 1; i++) {
			String s = br.readLine();
			String[] ss = s.trim().split(" ");
			int x = Integer.parseInt(ss[0]) - 1;
			int y = Integer.parseInt(ss[1]) - 1;
			arl[x].add(i);
			arl[y].add(i);
		}
		int flag = 0;
		int[] ans = new int[n - 1];
		HashSet<Integer> set = new HashSet<Integer>();
		for (int i = 0; i < n; i++) {
			if (arl[i].size() > 2) {
				flag = 1;
				ans[arl[i].get(0)] = 0;
				ans[arl[i].get(1)] = 1;
				ans[arl[i].get(2)] = 2;
				set.add(arl[i].get(0));
				set.add(arl[i].get(1));
				set.add(arl[i].get(2));
				break;
			}
		}
		if (flag == 0) {
			for (int i = 0; i < n - 1; i++) {
				System.out.println(i);
			}
		} else {
			int k = 3;
			for (int i = 0; i < n - 1; i++) {
				if (!set.contains(i))
					ans[i] = k++;
				System.out.println(ans[i]);
			}
		}
    }
}