#include <iostream>
#include <vector>
using namespace std;
int main(){
	//dp只能O(n3)，这里最好用bfs或者dfs 
    int v, e;
    cin >> v >> e;
    //vector<vector<int>> adj(v, vector<int>());
    vector<vector<bool>> dp(v, vector<bool>(v, false));
    for(int i = 0; i < e; i++){
        int a, b;
        cin >> a >> b;
        //adj[a - 1].push_back(b - 1);
        //adj[b - 1].push_back(a - 1);
        dp[a - 1][b - 1] = true;
    }
	//考虑dp[i][j] : 两个点是否连通，采取插入k的方式解决 
    for (int i = 0; i < v; i++) dp[i][i] = true; // 自己到自己连通
	for(int k = 0; k < v; k++){
		for(int i = 0; i < v; i++){
			for(int j = 0; j < v; j++){
				if(dp[i][j]) continue;
				dp[i][j] = dp[i][j] || (dp[i][k] & dp[k][j]);
			}
		}
	}
	
	//bfs
	for(int i = 0; i < v; i++) {
    vector<bool> visited(v, false);
    queue<int> q;
    q.push(i);
    visited[i] = true;
    int mx = i;
    while (!q.empty()) {
        int curr = q.front();
        q.pop();
        for (int next : adj[curr]) {
            if (!visited[next]) {
                visited[next] = true;
                q.push(next);
                mx = max(mx, next);
            }
        }
    }
    cout << mx + 1 << " ";
} 
	
	
	
	
	for(int i = 0; i < v; i++){
		int mx = -1;
		for(int j = 0; j < v; j++){
			if(dp[i][j] && mx < j) mx = j;
		}
		cout << mx + 1 << " ";
	}
	cout << endl;
    return 0;
}