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.*;
import java.util.*;
 
 
public class  Main{
	static LinkedList<int[]>[]adj;
	public static void main(String[] args) throws Exception{
		MScanner sc=new MScanner(System.in);
		PrintWriter pw=new PrintWriter(System.out);
		int n=sc.nextInt();
		adj=new LinkedList[n];
		for(int i=0;i<n;i++)adj[i]=new LinkedList<int[]>();
		for(int i=0;i<n-1;i++) {
			int x=sc.nextInt()-1,y=sc.nextInt()-1;
			adj[x].add(new int[]{y,i});adj[y].add(new int[]{x,i});
		}
		int max=0;
		for(int i=0;i<n;i++) {
			if(adj[i].size()>adj[max].size()) {
				max=i;
			}
		}
		int[]ans=new int[n-1];
		Arrays.fill(ans, -1);
		int val=0;
		for(int []j:adj[max]) {
			ans[j[1]]=val++;
		}
		for(int i=0;i<n-1;i++) {
			if(ans[i]==-1) {
				ans[i]=val++;
			}
		}
		for(int i=0;i<n-1;i++) {
			pw.println(ans[i]);
		}
		
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] takearr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] takearrl(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public Integer[] takearrobj(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] takearrlobj(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}