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)

Scanf.scanf "%d %d" (fun n m ->
    let nei = Array.make n [] in
    let rev = Array.make n 0 in
    for i = 1 to n - 1 + m do
        Scanf.scanf " %d %d" (fun a b ->
            let a = a - 1 in
            let b = b - 1 in
            nei.(a) <- b :: nei.(a);
            rev.(b) <- rev.(b) + 1
        )
    done;
    let rec loop i =
        if rev.(i) = 0 then i else loop (i + 1)
    in
    let root = loop 0 in
    let par = Array.make n 0 in

    let rec loop next = function
        | [] -> if next <> [] then loop [] next
        | hd :: tl ->
            let rec loop2 next = function
                | [] -> loop next tl
                | x :: xs ->
                        rev.(x) <- rev.(x) - 1;
                        if rev.(x) > 0 then loop2 next xs else (
                            par.(x) <- hd + 1;
                            loop2 (x :: next) xs
                        )
            in
            loop2 next nei.(hd)
    in
    loop [] [ root ];
    Array.iter (Printf.printf "%d\n") par
)