#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>P;
vector<int>node[100009];
int ans[100009],mx[100009],p[100009],t[100009];

int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<n+m-1;i++){
    int a,b;
    cin>>a>>b;
    node[a].push_back(b);
    p[b]++;
  }
  int s;
  for(int i=1;i<=n;i++){
    if(p[i]==0){
      s=i;
    }
  }

  queue<P>que;
  que.push(P(s,0));
  while(!que.empty()){
    P now_p=que.front();que.pop();
    int now=now_p.first;
    int cnt=now_p.second;
    for(int i=0;i<node[now].size();i++){
      int next=node[now][i];
      if(cnt+1>mx[next]){
	mx[next]=cnt+1;
	ans[next]=now;
      }
      t[next]++;
      if(t[next]==p[next]){
	que.push(P(next,cnt+1));
      }
    }
  }
  
  for(int i=1;i<=n;i++){
    cout<<ans[i]<<endl;
  }
  return(0);
}

### Prompt

Please formulate a CPP solution to the following problem:

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.

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.

Examples

Input

3 1
1 2
1 3
2 3


Output

0
1
2


Input

6 3
2 1
2 3
4 1
4 2
6 1
2 6
4 6
6 5


Output

6
4
2
0
6
2

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>P;
vector<int>node[100009];
int ans[100009],mx[100009],p[100009],t[100009];

int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<n+m-1;i++){
    int a,b;
    cin>>a>>b;
    node[a].push_back(b);
    p[b]++;
  }
  int s;
  for(int i=1;i<=n;i++){
    if(p[i]==0){
      s=i;
    }
  }

  queue<P>que;
  que.push(P(s,0));
  while(!que.empty()){
    P now_p=que.front();que.pop();
    int now=now_p.first;
    int cnt=now_p.second;
    for(int i=0;i<node[now].size();i++){
      int next=node[now][i];
      if(cnt+1>mx[next]){
	mx[next]=cnt+1;
	ans[next]=now;
      }
      t[next]++;
      if(t[next]==p[next]){
	que.push(P(next,cnt+1));
      }
    }
  }
  
  for(int i=1;i<=n;i++){
    cout<<ans[i]<<endl;
  }
  return(0);
}

```