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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        ArrayList<Edge>[] tree;
        int N;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            N = in.nextInt();
            tree = new ArrayList[N];
            if (N == 2) {
                out.println(0);
            } else {
                for (int i = 0; i < N; i++) {
                    tree[i] = new ArrayList<>();
                }
                for (int i = 0; i < N - 1; i++) {
                    int a = in.nextInt() - 1;
                    int b = in.nextInt() - 1;
                    tree[a].add(new Edge(b, i));
                    tree[b].add(new Edge(a, i));
                }
                int cnt = 0;
                int[] ans = new int[N - 1];
                Arrays.fill(ans, -1);
                for (int i = 0; i < N; i++) {
                    if (tree[i].size() == 1) {
                        ans[tree[i].get(0).i] = cnt++;
                    }
                }
                if (cnt <= 1) {
                    for (int j = 0; j < N - 1; j++) {
                        out.println(j);
                    }
                } else {
                    for (int i = 0; i < N - 1; i++) {
                        if (ans[i] == -1) {
                            out.println(cnt++);
                        } else {
                            out.println(ans[i]);
                        }
                    }
                }
            }
        }

        class Edge {
            int a;
            int i;

            Edge(int a, int i) {
                this.a = a;
                this.i = i;
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

