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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::collections::{BTreeSet, VecDeque};

fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { reader: s.lock() };
    let n: usize = sc.read();
    let m: usize = sc.read();
    let mut graph = vec![vec![]; n];
    let mut ins = vec![0; n];
    for _ in 0..(n - 1 + m) {
        let a = sc.read::<usize>() - 1;
        let b = sc.read::<usize>() - 1;
        graph[a].push(b);
        ins[b] += 1;
    }

    let mut root = n;
    for i in 0..n {
        if ins[i] == 0 {
            assert_eq!(root, n);
            root = i;
        }
    }
    assert!(root != n);

    let mut parents = vec![n; n];
    let mut q = VecDeque::new();
    let mut dist = vec![0; n];
    dist[root] = 0;
    q.push_back(root);
    while let Some(v) = q.pop_front() {
        for &next in graph[v].iter() {
            if dist[next] < dist[v] + 1 {
                dist[next] = dist[v] + 1;
                parents[next] = v;
                q.push_back(next);
            }
        }
    }
    for i in 0..n {
        if i != root {
            println!("{}", parents[i] + 1);
        } else {
            println!("0");
        }
    }
}

pub struct Scanner<R> {
    reader: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .reader
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n')
            .take_while(|&b| b != b' ' && b != b'\n')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}