Problem:
Score : 500 points

Problem Statement

There is a rooted tree (see Notes) with N vertices numbered 1 to N.
Each of the vertices, except the root, has a directed edge coming from its parent.
Note that the root may not be Vertex 1.

Takahashi has added M new directed edges to this graph.
Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v.

You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges.
More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i.

Restore the original rooted tree.

Notes

For "tree" and other related terms in graph theory, see the article in Wikipedia, for example.

Constraints

3 \leq N

1 \leq M

N + M \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

If i \neq j, (A_i, B_i) \neq (A_j, B_j).

The graph in input can be obtained by adding M edges satisfying the condition in the problem statement to a rooted tree with N vertices.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_{N-1+M} B_{N-1+M}

Output

Print N lines.
In the i-th line, print 0 if Vertex i is the root of the original tree, and otherwise print the integer representing the parent of Vertex i in the original tree.

Note that it can be shown that the original tree is uniquely determined.

Sample Input 1

3 1
1 2
1 3
2 3

Sample Output 1

0
1
2

The graph in this input is shown below:

It can be seen that this graph is obtained by adding the edge 1 \rightarrow 3 to the rooted tree 1 \rightarrow 2 \rightarrow 3.

Sample Input 2

6 3
2 1
2 3
4 1
4 2
6 1
2 6
4 6
6 5

Sample Output 2

6
4
2
0
6
2

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using System.Numerics;

namespace NikkeiContest.D
{
	using static Util;

	public class Solver : SolverBase
	{
		public void Run() {
			var ary = ReadIntArray();
			var N = ary[0];
			var M = ary[1];

			// [i] = 当該頂点への入次数（頂点に入ってくる数）
			int[] indegrees = new int[N+1];
			// tree[i] = 子供のList
			var tree = new List<int>[N+1];
			for (int i = 0; i < N + M - 1; i++) {
				var a = ReadIntArray();
				var p = a[0];
				var c = a[1];

				indegrees[c]++;

				var l = tree[p];
				if (l == null) tree[p] = l = new List<int>();
				l.Add(c);
			}

			// root親を探す ※入次数=0 がroot
			int root = -1;
			for (int i = 1; i <= N; i++) {
				if (indegrees[i] == 0) {
					root = i;
				}
			}

			// aryP[i] = 頂点i の 親番号
			var aryP = new int[N + 1];

			// BFSで子供をたどる
			var que = new Queue<int>();
			que.Enqueue(root);

			// BFSでトポロジカルソートしながら、親を確定していく
			while (0 < que.Count) {
				var p = que.Dequeue();
				if (tree[p] == null) continue;

				foreach (int c in tree[p]) {
					// 子でループ
					// 子の入次数をへらす
					indegrees[c]--;

					// 子の入次数が0 ＝ この時点で親が確定
					if (indegrees[c] == 0) {
						que.Enqueue(c);
						aryP[c] = p;
					}
				}
			}
			for (int i = 1; i < N + 1; i++) {
				WriteLine(aryP[i]);
			}
		}

#if !MYHOME
		public static void Main(string[] args) {
			new Solver().Run();
		}
#endif
	}

	public static class Util
	{
		public readonly static long MOD = 1000000007;

		public static string DumpToString<T>(IEnumerable<T> array) where T : IFormattable {
			var sb = new StringBuilder();
			foreach (var item in array) {
				sb.Append(item);
				sb.Append(", ");
			}
			return sb.ToString();
		}

		public static void InitArray<T>(T[] ary, T value) {
			for (int i = 0; i < ary.Length; i++) {
				ary[i] = value;
			}
		}
		public static void InitDP2<T>(T[,] dp, T value) {
			for (int i = 0; i < dp.GetLength(0); i++) {
				for (int j = 0; j < dp.GetLength(1); j++) {
					dp[i, j] = value;
				}
			}
		}
		public static void InitDP3<T>(T[,,] dp, T value) {
			for (int i = 0; i < dp.GetLength(0); i++) {
				for (int j = 0; j < dp.GetLength(1); j++) {
					for (int k = 0; k < dp.GetLength(2); k++) {
						dp[i, j, k] = value;
					}
				}
			}
		}

		public static T Max<T>(params T[] nums) where T : IComparable {
			if (nums.Length == 0) return default(T);

			T max = nums[0];
			for (int i = 1; i < nums.Length; i++) {
				max = max.CompareTo(nums[i]) > 0 ? max : nums[i];
			}
			return max;
		}
		public static T Min<T>(params T[] nums) where T : IComparable {
			if (nums.Length == 0) return default(T);

			T min = nums[0];
			for (int i = 1; i < nums.Length; i++) {
				min = min.CompareTo(nums[i]) < 0 ? min : nums[i];
			}
			return min;
		}

		/// <summary>
		/// ソート済み配列 ary に同じ値の要素が含まれている？
		/// ※ソート順は昇順/降順どちらでもよい
		/// </summary>
		public static bool HasDuplicateInSortedArray<T>(T[] ary) where T : IComparable, IComparable<T> {
			if (ary.Length <= 1) return false;

			var lastNum = ary[ary.Length - 1];

			foreach (var n in ary) {
				if (lastNum.CompareTo(n) == 0) {
					return true;
				} else {
					lastNum = n;
				}
			}
			return false;
		}

		/// <summary>
		/// 二分探索
		/// ※条件を満たす最小のidxを返す
		/// ※満たすものがない場合は ary.Length を返す
		/// ※『aryの先頭側が条件を満たさない、末尾側が条件を満たす』という前提
		/// ただし、IsOK(...)の戻り値を逆転させれば、逆でも同じことが可能。
		/// </summary>
		/// <param name="ary">探索対象配列 ★ソート済みであること</param>
		/// <param name="key">探索値 ※これ以上の値を持つ（IsOKがtrueを返す）最小のindexを返す</param>
		public static int BinarySearch<T>(T[] ary, T key) where T : IComparable, IComparable<T> {
			int left = -1;
			int right = ary.Length;

			while (1 < right - left) {
				var mid = left + (right - left) / 2;

				if (IsOK(ary, mid, key)) {
					right = mid;
				} else {
					left = mid;
				}
			}

			// left は条件を満たさない最大の値、right は条件を満たす最小の値になっている
			return right;
		}
		public static bool IsOK<T>(T[] ary, int idx, T key) where T : IComparable, IComparable<T> {
			// key <= ary[idx] と同じ意味
			return key.CompareTo(ary[idx]) <= 0;
		}
	}

	public class SolverBase
	{
		virtual protected string ReadLine() => Console.ReadLine();
		virtual protected int ReadInt() => int.Parse(ReadLine());
		virtual protected long ReadLong() => long.Parse(ReadLine());
		virtual protected string[] ReadStringArray() => ReadLine().Split(' ');
		virtual protected int[] ReadIntArray() => ReadLine().Split(' ').Select<string, int>(s => int.Parse(s)).ToArray();
		virtual protected long[] ReadLongArray() => ReadLine().Split(' ').Select<string, long>(s => long.Parse(s)).ToArray();
		virtual protected double[] ReadDoubleArray() => ReadLine().Split(' ').Select<string, double>(s => double.Parse(s)).ToArray();
		virtual protected void WriteLine(string line) => Console.WriteLine(line);
		virtual protected void WriteLine(double d) => Console.WriteLine($"{d:F9}");
		virtual protected void WriteLine<T>(T value) where T : IFormattable => Console.WriteLine(value);

		[Conditional("DEBUG")]
		protected void Dump(string s) => Console.WriteLine(s);
		[Conditional("DEBUG")]
		protected void Dump(char c) => Console.WriteLine(c);
		[Conditional("DEBUG")]
		protected void Dump(double d) => Console.WriteLine($"{d:F9}");
		[Conditional("DEBUG")]
		protected void Dump<T>(IEnumerable<T> array) where T : IFormattable {
			string s = Util.DumpToString(array);
			// Consoleに出力すると、UnitTestの邪魔をしないというメリットあり。
			Console.WriteLine(s);
			//_writer.WriteLine(s);
		}
		[Conditional("DEBUG")]
		protected void DumpGrid<T>(IEnumerable<IEnumerable<T>> arrayOfArray) where T : IFormattable {
			var sb = new StringBuilder();
			foreach (var a in arrayOfArray) {
				sb.AppendLine(Util.DumpToString(a));
			}
			// Consoleに出力すると、UnitTestの邪魔をしないというメリットあり。
			Console.WriteLine(sb.ToString());
			//_writer.WriteLine(sb.ToString());
		}
		[Conditional("DEBUG")]
		protected void DumpDP<T>(T[,] dp) where T : IFormattable {
			var sb = new StringBuilder();
			for (int i = 0; i < dp.GetLength(0); i++) {
				for (int j = 0; j < dp.GetLength(1); j++) {
					sb.Append(dp[i, j]);
					sb.Append(", ");
				}
				sb.AppendLine();
			}
			Console.WriteLine(sb.ToString());
		}
	}
}