Problem:
Score : 1200 points

Problem Statement

Takahashi's office has N rooms. Each room has an ID from 1 to N.
There are also N-1 corridors, and the i-th corridor connects Room a_i and Room b_i.
It is known that we can travel between any two rooms using only these corridors.

Takahashi has got lost in one of the rooms. Let this room be r.
He decides to get back to his room, Room 1, by repeatedly traveling in the following manner:

Travel to the room with the smallest ID among the rooms that are adjacent to the rooms already visited, but not visited yet.

Let c_r be the number of travels required to get back to Room 1.
Find all of c_2,c_3,...,c_N.
Note that, no matter how many corridors he passes through in a travel, it still counts as one travel.

Constraints

2 \leq N \leq 2\times 10^5

1 \leq a_i,b_i \leq N

a_i \neq b_i

The graph given as input is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

Print c_r for each r, in the following format:

c_2 c_3 ... c_N

Sample Input 1

6
1 5
5 6
6 2
6 3
6 4

Sample Output 1

5 5 5 1 5

For example, if Takahashi was lost in Room 2, he will travel as follows:

Travel to Room 6.

Travel to Room 3.

Travel to Room 4.

Travel to Room 5.

Travel to Room 1.

Sample Input 2

6
1 2
2 3
3 4
4 5
5 6

Sample Output 2

1 2 3 4 5

Sample Input 3

10
1 5
5 6
6 10
6 4
10 3
10 8
8 2
4 7
4 9

Sample Output 3

7 5 3 1 3 4 7 4 5

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Diagnostics;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using static System.Math;
using Number = System.Int32;

using C = System.Int32;
using V = System.Int64;
namespace Program {
	public class Solver {
		Random rnd = new Random();
		public void Solve() {
			var n = ri;
			var G = Enumerate(n, x => new List<int>());
			for (int i = 0; i < n - 1; i++) {
				var u = ri - 1;
				var v = ri - 1;
				G[u].Add(v);
				G[v].Add(u);
			}
			var par = Enumerate(n, x => -50);
			par[0] = -1;
			var d = new int[n];
			d[0] = 0;
			var mi = Enumerate(n, x => x);
			var q = new Queue<int>(); q.Enqueue(0);
			mi[0] = 0;
			var S = Enumerate(n, x => new List<int>());
			while (q.Any()) {
				var p = q.Dequeue();

				foreach (var t in G[p])
					if (par[t] == -50) {
						par[t] = p; q.Enqueue(t);
						mi[t] = Min(p, mi[p]);
						S[mi[t]].Add(t);
						d[t] = d[p] + 1;
					}
			}
			var s = new DisjointSet(n);
			var val = new int[n];
			for (int i = 0; i < n; i++) {
				foreach (var v in S[i])
					foreach (var c in G[v])
						if (c != par[v] && c < par[v]) {
							val[v] += s.Size(c);
							val[c] -= s.Size(c);
						}
				foreach (var c in G[i])
					if (c != par[i] && c < i) s.Unite(i, c);
			}
			q.Enqueue(0);
			while (q.Any()) {
				var p = q.Dequeue();
				foreach (var t in G[p])
					if (t != par[p]) {
						val[t] += val[p];
						q.Enqueue(t);
					}
			}
			for (int i = 0; i < n; i++) val[i] += d[i];
			Console.WriteLine(val.Skip(1).AsJoinedString());
		}

		const long INF = 1L << 60;
		static int[] dx = { -1, 0, 1, 0 };
		static int[] dy = { 0, 1, 0, -1 };
		int ri { get { return sc.Integer(); } }
		long rl { get { return sc.Long(); } }
		double rd { get { return sc.Double(); } }
		string rs { get { return sc.Scan(); } }
		public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

		static T[] Enumerate<T>(int n, Func<int, T> f) {
			var a = new T[n];
			for (int i = 0; i < a.Length; ++i) a[i] = f(i);
			return a;
		}
		static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
	}
}

#region main
static class Ex {
	static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
	static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
		return string.Join(st, ie);
	}
	static public void Main() {
		Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = true });
		var solver = new Program.Solver();
		//* 
		var t = new System.Threading.Thread(solver.Solve, 50000000);
		t.Start();
		t.Join();
		//*/
		// solver.Solve();
		Console.Out.Flush();
	}
}
#endregion
#region Ex
namespace Program.IO {
	using System.IO;
	using System.Text;
	using System.Globalization;

	public class Printer : StreamWriter {
		public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
		public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
	}

	public class StreamScanner {
		public StreamScanner(Stream stream) { str = stream; }

		public readonly Stream str;
		private readonly byte[] buf = new byte[1024];
		private int len, ptr;
		public bool isEof = false;
		public bool IsEndOfStream { get { return isEof; } }

		private byte read() {
			if (isEof) return 0;
			if (ptr >= len) {
				ptr = 0;
				if ((len = str.Read(buf, 0, 1024)) <= 0) {
					isEof = true;
					return 0;
				}
			}
			return buf[ptr++];
		}

		public char Char() {
			byte b = 0;
			do b = read(); while ((b < 33 || 126 < b) && !isEof);
			return (char)b;
		}
		public string Scan() {
			var sb = new StringBuilder();
			for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
			return sb.ToString();
		}
		public string ScanLine() {
			var sb = new StringBuilder();
			for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
			return sb.ToString();
		}
		public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
		public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
		public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
	}
}

#endregion

#region Disjoint Set
/// <summary>
/// いくつかの要素を素集合に分割して保持するコレクションです．
/// </summary>
public class DisjointSet {
	int[] par;
	byte[] rank;
	/// <summary>
	/// 0 から <paramref name="N"/>-1 までの番号がついた <paramref name="N"/> 個の要素からなる集合を作成します．
	/// </summary>
	/// <param name="N">要素数</param>
	/// <remarks>このコンストラクタは O(N) で実行されます．</remarks>
	public DisjointSet(int N) {
		par = new int[N];
		rank = new byte[N];
		for (int i = 0; i < N; i++)
			par[i] = -1;
	}
	/// <summary>
	/// 指定した要素が属する集合の代表値を取得します．
	/// </summary>
	/// <param name="id">調べたい要素の 0-indexed での番号</param>
	/// <returns>指定した要素が属する集合の代表値</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public int this[int id] {
		get {
			if ((par[id] < 0)) return id;
			return par[id] = this[par[id]];
		}
	}
	/// <summary>
	/// 指定した 2 つの要素が属する集合同士を 1 つに統合することを試みます．
	/// </summary>
	/// <param name="x">最初の要素の 0-indexed での番号</param>
	/// <param name="y">2 つ目の要素の 0-indexed での番号</param>
	/// <returns>統合に成功したならば true，そうでなければ false．</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public bool Unite(int x, int y) {
		x = this[x]; y = this[y];
		if (x == y) return false;
		if (rank[x] < rank[y]) { var tmp = x; x = y; y = tmp; }
		par[x] += par[y];
		par[y] = x;
		if (rank[x] == rank[y])
			rank[x]++;
		return true;
	}
	/// <summary>
	/// 指定した要素が属する集合のサイズを取得します．
	/// </summary>
	/// <param name="x">指定する要素の 0-indexed での番号</param>
	/// <returns>集合のサイズ</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public int Size(int x) { return -par[this[x]]; }

	/// <summary>
	/// 指定した 2 つの要素が属する集合が同じかどうかを判定する．計算量 O(α(N))．
	/// </summary>
	/// <param name="x">最初の要素の 0-indexed での番号</param>
	/// <param name="y">2 つ目の要素の 0-indexed での番号</param>
	/// <returns>同じならば true，そうでなければ false．</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public bool Same(int x, int y) { return this[x] == this[y]; }

}
#endregion