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 Python (3.4.3) (judge verdict: Runtime Error)

# cook your dish here
n,m = map(int,input().split())

par = [[] for i in range(n)]
op = [[]for i in range(n)]
for i in range(n+m-1):
    x,y = map(int,input().split())
    x-=1
    y-=1
    par[y].append(x)
    op[x].append(y)

res = [-1]*n
x = -1


for i in range(n):
    if len(par[i])==0:
        res[i] = 0
        x = i
        break


used = [0]*n
used[x]=1

#不使用の親がいるか
def g(y):
    for pe in par[y]:
        if used[pe]==0:
            return False
    return True
    
#xの下のusedとresを埋める
def f(z):
    for e in op[z]:
        if used[e]==0 and g(e):
                used[e]=1
                #res[e]=z+1
                f(e)

f(x)

for i in range(n):
    print(res[i])