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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"bytes"
	"fmt"
	"math"
	"os"
	"strconv"
)

var edges, redges map[int][]int
var memo map[int]int

func main() {
	scanner := bufio.NewScanner(os.Stdin)
	var N, M int
	scanner.Scan()
	fmt.Sscanf(scanner.Text(), "%d %d", &N, &M)

	edges = make(map[int][]int)
	redges = make(map[int][]int)
	memo = make(map[int]int)

	as := make([]int, N+M-1)
	bs := make([]int, N+M-1)

	for i := 0; i < N+M-1; i++ {
		scanner.Scan()
		fmt.Sscanf(scanner.Text(), "%d %d", &as[i], &bs[i])

		cs, e := edges[as[i]]
		if !e {
			cs = make([]int, 0)
		}
		edges[as[i]] = append(cs, bs[i])

		ps, e := redges[bs[i]]
		if !e {
			ps = make([]int, 0)
		}
		redges[bs[i]] = append(ps, as[i])
	}

	root := -1
	for i := 1; i <= N; i++ {
		if _, e := redges[i]; !e {
			root = i
			break
		}
	}
	//fmt.Println(root)
	//fmt.Println(reachables)

	orgrEdge := make(map[int]int)
	for target, parents := range redges {
		if len(parents) >= 2 {
			tmpLen := math.MaxInt64
			tmpParent := -1
			for _, parent := range redges[target] {
				l := reachables(parent)
				if l < tmpLen {
					tmpLen = l
					tmpParent = parent
				}
			}
			orgrEdge[target] = tmpParent
		}
	}

	var buffer bytes.Buffer
	for i := 1; i <= N; i++ {
		if i == root {
			buffer.WriteString("0\n")
		} else if from, e := orgrEdge[i]; e {
			buffer.WriteString(strconv.Itoa(from))
			buffer.WriteString("\n")
			continue
		} else {
			froms := redges[i]
			buffer.WriteString(strconv.Itoa(froms[0]))
			buffer.WriteString("\n")
		}
	}
	fmt.Println(buffer.String())

}

func reachables(currentNode int) int {
	if nodes, e := memo[currentNode]; e {
		return nodes
	}
	ret := 0
	ret++
	for _, child := range edges[currentNode] {
		ret = ret + reachables(child)
	}
	memo[currentNode] = ret
	return ret
}