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.io.*;
import java.util.*;

public class Main {

	public static void main(String[] args) throws Exception {
		
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		
		int n = Integer.parseInt(in.readLine());
		
		int[] v = new int[n+1];
		
		int node = 0;
		boolean done = false;
		
		int[][] edge = new int[n][2];
		for(int i = 0; i < n-1; i++) {
			StringTokenizer st = new StringTokenizer(in.readLine());
			int a = Integer.parseInt(st.nextToken());
			int b = Integer.parseInt(st.nextToken());
			v[a]++;
			v[b]++;
			
			edge[i] = new int[] {a, b};
			if(v[a] >= 3 && !done) {
				node = a;
				done = true;
			}
			
			if(v[b] >= 3 && !done) {
				node = b;
				done = true;
			}
		}
		
		int count = 0;
		int count2 = 3;
		for(int i = 0; i < n-1; i++) {
			if(done) {
				if((edge[i][0] == node || edge[i][1] == node) && count < 3) {
					System.out.println(count);
					count++;
				} else {
					System.out.println(count2);
					count2++;
				}
			} else {
				System.out.println(count);
				count++;
			}
		}
	}
}
