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 IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int n = Integer.parseInt(br.readLine());
        ArrayList<Integer>[] adj = new ArrayList[n];
        for(int i = 0; i<n; i++) adj[i] = new ArrayList<>();
        for(int i = 0; i<n-1; i++){
            String[] str = br.readLine().split(" ");
            int u = Integer.parseInt(str[0])-1, v = Integer.parseInt(str[1])-1;
            adj[u].add(i);
            adj[v].add(i);
        }
        int[] ans = new int[n-1];
        Arrays.fill(ans, -1);
        int num = 0;
        for(int i = 0; i<n; i++){
            if(adj[i].size()>=3){
                for(int j = 0; j<3; j++){
                    ans[adj[i].get(j)] = num++;
                }
                break;
            }
        }
        for(int i = 0; i<n-1; i++){
            if(ans[i] == -1) ans[i] = num++;
        }
        for(int i = 0; i<n-1; i++){
            System.out.println(ans[i]);
        }
    }
}