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 test {
    int INF = (int)1e9;
    long MOD = 1000000007;
    

    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int[][] e = new int[n-1][2];
        int[] deg = new int[n];
        for(int i=0; i<n-1; i++) {
            e[i][0] = in.nextInt()-1;
            e[i][1] = in.nextInt()-1;
            deg[e[i][0]]++;
            deg[e[i][1]]++;
        }
        int c = 0;
        for(int i=0; i<n; i++) {
            if(deg[i]==1) c++;
        }
        if(c>=3) {
            int[] o = new int[n-1];
            int x = 1;
            loop1: for(int i=0; i<n-1; i++) {
                for(int j=0; j<=1; j++) {
                    if(deg[e[i][j]]==1) {
                        o[i] = x++;
                        deg[e[i][j]]++;
                    }
                    if(x>3) {
                        break loop1;
                    }
                }
            }
            x--;
            for(int i=0; i<n-1; i++) {
                if(o[i]>0) {
                    out.println(o[i]-1);
                } else {
                    out.println(x++);
                }
            }
        } else {
            for(int i=0; i<n-1; i++) {
                out.println(i);
            }
        }
    }
    

    public static void main(String[] args) throws IOException {
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        while(t-- >0) {
            new test().solve(in, out);
        }
        out.close();
    }
    
    public static void show(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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());
        }
    }
}