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 C 
{	
	
	static ArrayList<Integer> g[];
	
	public static void main(String[] args)throws IOException 
	{	
		FastReader f=new FastReader();
		StringBuffer sb = new StringBuffer();
		
		int n=f.nextInt();
		
		g=new ArrayList[n+1];
		for(int i=0;i<=n;i++)
			g[i]=new ArrayList<Integer>();
		
		int arr[][]=new int[n+1][2];
		for(int i=1;i<n;i++)
		{
			int x=f.nextInt();
			int y=f.nextInt();
			arr[i][0]=x;
			arr[i][1]=y;
			
			g[x].add(y);
			g[y].add(x);	
		}
		
		boolean ok=false;
		int x,y1,y2,y3;
		x=y1=y2=y3=0;
		for(int i=1;i<=n;i++)
		{
			if(g[i].size()>=3)
			{
				ok=true;
				x=i;
				y1=g[i].get(0);
				y2=g[i].get(1);
				y3=g[i].get(2);
				
				break;
			}
		}
		
//		System.out.println("ok ? = "+ok);
		
		String so1=x+" "+y1;
		String so2=x+" "+y2;
		String so3=x+" "+y3;
//		System.out.println("so1 = "+so1);
//		System.out.println("so2 = "+so2);
//		System.out.println("so3 = "+so3);

		
		if(ok)
		{
			int curr=3;
			for(int i=1;i<n;i++)
			{	
				String s1=arr[i][0]+" "+arr[i][1];
				String s2=arr[i][1]+" "+arr[i][0];
				
//				System.out.println("s1 = "+s1);
//				System.out.println("s2 = "+s2);
				
				if(s1.equals(so1) || s2.equals(so1))
					sb.append("0\n");
				else if(s1.equals(so2) || s2.equals(so2))
					sb.append("1\n");
				else if(s1.equals(so3) || s2.equals(so3))
					sb.append("2\n");
				else
				{
					sb.append(curr+"\n");
					curr++;
				}
			}
		}
		else
		{
			int curr=0;
			for(int i=1;i<n;i++)
			{
				sb.append(curr+"\n");
				curr++;
			}
		}
		System.out.println(sb);
	}

	static class FastReader 
	{ 
	    BufferedReader br; 
	    StringTokenizer st; 
 
	    public FastReader() {
	    	br = new BufferedReader(new
	                 InputStreamReader(System.in)); 
	    }
	    String next() { 
	        while (st == null || !st.hasMoreElements()) { 
	            try{ 
	                st = new StringTokenizer(br.readLine()); 
	            } 
	            catch (IOException  e) { 
	                e.printStackTrace(); 
	            } 
	        } 
	        return st.nextToken(); 
	    } 
	    int nextInt() { 
	        return Integer.parseInt(next()); 
	    } 
	    long nextLong() { 
	        return Long.parseLong(next()); 
	    } 
	    double nextDouble() { 
	        return Double.parseDouble(next()); 
	    } 
	    String nextLine() { 
	        String str = ""; 
	        try{ 
	            str = br.readLine(); 
	        } 
	        catch (IOException e) { 
	            e.printStackTrace(); 
	        } 
	        return str; 
	    } 
	}
}