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 OCaml (4.02.3) (judge verdict: Accepted)

let rec fix f x = f (fix f) x;;
Scanf.(Array.(scanf " %d %d" @@ fun n m ->
  let e = make n [] in
  let e_rev = make n [] in
  init (n-1+m) (fun i -> scanf " %d %d" @@ fun u v ->
    let u,v = u-1,v-1 in
    e.(u) <- v :: e.(u);
    e_rev.(v) <- u :: e_rev.(v)
  ) |> ignore;
  let root = fst @@ fold_left (fun (root,i) ls ->
    (if List.length ls = 0 then i else root),i+1
  ) (0,0) e_rev in
  let q = Queue.create () in
  let v_left = init n (fun i -> List.length e_rev.(i)) in
  Queue.add root q;
  let par = make n 0 in
  while not @@ Queue.is_empty q do
    let i = Queue.pop q in
    List.iter (fun j ->
      v_left.(j) <- v_left.(j) - 1;
      if v_left.(j) = 0 then (
        par.(j) <- i + 1;
        Queue.add j q;
      )
    ) e.(i)
  done;
  iter (Printf.printf "%d\n") par
))