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
{
	
	private void solve()throws IOException
	{
		int n=nextInt();
		int deg[]=new int[n+1];
		int label[]=new int[n];
		int edge[]=new int[n+1];
		Arrays.fill(label,-1);
		for(int i=1;i<n;i++)
		{
			int u=nextInt();
			int v=nextInt();
			deg[u]++;
			deg[v]++;
			edge[u]=i;
			edge[v]=i;
		}
		if(n==2)
		{
			out.println(0);
			return;
		}
		int count=0;
		for(int i=1;i<=n;i++)
		{
			if(deg[i]==1)
			{
				label[edge[i]]=count;
				count++;
			}
			if(count==3)
				break;
		}
		for(int i=1;i<n;i++)
		{
			if(label[i]==-1)
				label[i]=count++;
		}
		for(int i=1;i<n;i++)
			out.println(label[i]);
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws IOException
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		solve();

		br.close();
		out.close();
	}
	public static void main(String args[])throws IOException{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws IOException{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws IOException{
		return br.readLine();
	}
	int nextInt()throws IOException{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws IOException{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws IOException{
		return Double.parseDouble(nextToken());
	}
}