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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int main(void){
    int n,m,i,r,f=1;
    scanf("%d%d",&n,&m);
    int a[n+m-1],b[n+m-1],noto[n],d[n],o[n];
    for(i=0;i<n;i++){noto[i]=0;o[i]=0;}
    for(i=0;i<n+m-1;i++){scanf("%d%d",&a[i],&b[i]);noto[b[i]-1]=1;}
    for(i=0;i<n;i++)if(!noto[i])r=i;
    for(i=0;i<n;i++){
        if(i==r)d[i]=0;
        else d[i]=-1000000;
    }
    while(f){
        f=0;
        for(i=0;i<n+m-1;i++){
            if(d[a[i]-1]+1>d[b[i]-1]){
                d[b[i]-1]=d[a[i]-1]+1;
                f|=1;
            }
        }
    }
    for(i=0;i<n+m-1;i++){
        if(d[a[i]-1]+1==d[b[i]-1])o[b[i]-1]=a[i];
    }
    for(i=0;i<n;i++){
        printf("%d\n",o[i]);
    }
}