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 tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static long mod = 1000000007;
	static int inf = (int) 1e15;
	static ArrayList<Integer>[] ad;
	static int n;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		n = sc.nextInt();
		ad = new ArrayList[n];
		for (int i = 0; i < n; i++)
			ad[i] = new ArrayList<>();
		int[] ans = new int[n - 1];
		TreeMap<seg,Integer> t = new TreeMap<>();
		for (int i = 0; i < n - 1; i++) {
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			ad[a].add(b);
			ad[b].add(a);
			t.put(new seg(a, b, 0),i);
		}
		Arrays.fill(ans, -1);
		boolean f = false;
		for (int i = 0; i < n; i++)
			if (ad[i].size() >= 3) {
				int k1 = ad[i].get(0);
				int k2 = ad[i].get(1);
				int k3 = ad[i].get(2);
				f = true;
				if (t.containsKey(new seg(i, k1, 0))) {
                          ans[t.get(new seg(i, k1, 0))]=0;
				} else {
					  ans[t.get(new seg(k1, i, 0))]=0;
				}
				if (t.containsKey(new seg(i, k2, 0))) {
					 ans[t.get(new seg(i, k2, 0))]=1;
				} else {
					 ans[t.get(new seg(k2, i, 0))]=1;
				}
				if (t.containsKey(new seg(i, k3, 0))) {
					 ans[t.get(new seg(i, k3, 0))]=2;
				} else {
					 ans[t.get(new seg(k3, i, 0))]=2;
				}
				break;
			}
		if(f) {
			int c=3;
			for(int i=0;i<n-1;i++)
				if(ans[i]==-1)
					ans[i]=c++;
		}
		else {
			int c=0;
			for(int i=0;i<n-1;i++)
				if(ans[i]==-1)
					ans[i]=c++;
		}
		for (int i = 0; i < n - 1; i++)
			out.println(ans[i]);
		out.flush();
	}

	static class seg implements Comparable<seg> {
		int a;
		int b;
		int id;

		seg(int aa, int bb, int vv) {
			a = aa;
			b = bb;
			id = vv;
		}

		public String toString() {
			return a + " " + b + " " + id;
		}

		public int compareTo(seg o) {
			if (a == o.a)
				return b - o.b;
			return a - o.a;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}