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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_ON(bit, i) (bit & (1LL << i))
#define BIT_COUNT(bit) (__builtin_popcount(bit))

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;
template<typename T> using VVVV = std::vector<std::vector<std::vector<std::vector<T>>>>;

template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline void print_2vec(const std::vector<std::vector<T>> &v) { for (int i = 0; i < v.size(); ++i) { for (int j = 0; j < v[0].size(); ++j) { if (j != 0) { std::cout << " "; } std::cout << std::setw(13) << std::left << v[i][j]; } std::cout << "\n";}}
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<typename T> T &chmin(T &a, const T &b) { return a = std::min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = std::max(a, b); }
template<class T> inline std::vector<T> unique(std::vector<T> v) { sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end()), v.end()); return v; }
// 初項s, 交差dのn個の数列の和
long long sum_of_arithmetic_progression(long long s, long long d, long long n) { return n * (2 * s + (n - 1) * d) / 2; }

const int INF = 1 << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = {0, -1, 0, 1}, dx4 = {1, 0, -1, 0};
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };

using namespace std;


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

    int N, M;
    cin >> N >> M;
    V<V<int>> out(N + 1);
    V<V<int>> in(N + 1);
    FOR(i, 0, N - 1 + M) {
        int a, b;
        cin >> a >> b;
        out[a].emplace_back(b);
        in[b].emplace_back(a);
    }

    int root = -1;
    FOR(i, 1, N + 1) {
        if (in[i].empty()) {
            root = i;
        }
    }

    V<int> ans(N + 1, -1);
    queue<int> que;
    que.emplace(root);

    while (not que.empty()) {
        int node = que.front(); que.pop();

        FOE(child, out[node]) {
            que.emplace(child);
            ans[child] = node;
        }
    }

    ans[root] = 0;
    FOR(i, 1, N + 1) {
        print(ans[i]);
    }

	return 0;
}