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

public class Solution{
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out=new PrintWriter(System.out);
		String[] temp=br.readLine().trim().split(" ");
		int n=Integer.parseInt(temp[0]);
		int[][] edges=new int[n-1][2];
		int[] degree=new int[n+1];
		for(int i=0;i<n-1;i++){
		    temp=br.readLine().trim().split(" ");
		    int u=Integer.parseInt(temp[0]);
		    int v=Integer.parseInt(temp[1]);
		    degree[u]++;
		    degree[v]++;
		    edges[i][0]=u;
		    edges[i][1]=v;
		}
		int count=0;
		int[] ans=new int[n-1];
		for(int i=0;i<n-1;i++){
		    int u=edges[i][0];
		    int v=edges[i][1];
		    if(degree[u]==1 || degree[v]==1)
		    {
		        ans[i]=count;
		        count++;
		    }
		}
		for(int i=0;i<n-1;i++){
		    int u=edges[i][0];
		    int v=edges[i][1];
		    if(degree[u]>1 && degree[v]>1){
		        ans[i]=count;
		        count++;
		    }
		}
		for(int i=0;i<n-1;i++){
		    out.println(ans[i]);
		}
		out.flush();
		out.close();
	}
}