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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class C {
    final int INF = 0x3f3f3f3f;
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    public C() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) {
        new C();
    }

    private void solve() {
        int n = nextInt();
        List<Set<Integer>> g = new ArrayList<>(n + 1);
        for (int i = 0; i <= n; i++) {
            g.add(new HashSet<>());
        }
        for (int i = 1; i < n; i++) {
            int from = nextInt();
            int to = nextInt();
            g.get(from).add(i);
            g.get(to).add(i);
        }
        int ind = 0;
        int max = 0;
        for (int i = 1; i <= n; i++) {
            if (max < g.get(i).size()) {
                max = g.get(i).size();
                ind = i;
            }
        }
        int cur = 0;
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        for (Integer i : g.get(ind)) {
            ans[i] = cur++;
        }
        for (int i = 1; i < n; i++) {
            if (ans[i] == -1) ans[i] = cur++;
            outln(ans[i]);
        }
    }


    private void outln(Object o) {
        out.println(o);
    }

    private void out(Object o) {
        out.print(o);
    }

    public long[] nextLongArr(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public int[] nextIntArr(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

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

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() {
        return Double.parseDouble(nextToken());
    }


}