#include<bits/stdc++.h>
#define ll long long
#define f first
#define s second
#define pb push_back
using namespace std;
ll n,m,p[200005],xx,yy,root,raod[200005],fix[200005];
vector<ll>v[200005];

void go(ll x,ll par){
	p[x] = par;
	fix[x] = 1;
	for(int i=0; i<v[x].size(); i++)
		raod[v[x][i]]--;
	
	for(int i=0; i<v[x].size(); i++){
		if(!raod[v[x][i]] && !fix[v[x][i]]){
			go(v[x][i],x);
		}
	}
}
int main(){
	cin >> n >> m;
	
	for(int i=1; i<=n+m-1; i++){
		cin >> xx >> yy;
		raod[yy]++;
		v[xx].pb(yy);
	}
	
	for(int i=1; i<=n; i++){
		if(!raod[i])root = i;
	}
	
	go(root,0);
	
	for(int i=1; i<=n; i++)cout << p[i] << endl;
	
	
	return 0;
}
### Prompt

Your challenge is to write 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>
#define ll long long
#define f first
#define s second
#define pb push_back
using namespace std;
ll n,m,p[200005],xx,yy,root,raod[200005],fix[200005];
vector<ll>v[200005];

void go(ll x,ll par){
	p[x] = par;
	fix[x] = 1;
	for(int i=0; i<v[x].size(); i++)
		raod[v[x][i]]--;
	
	for(int i=0; i<v[x].size(); i++){
		if(!raod[v[x][i]] && !fix[v[x][i]]){
			go(v[x][i],x);
		}
	}
}
int main(){
	cin >> n >> m;
	
	for(int i=1; i<=n+m-1; i++){
		cin >> xx >> yy;
		raod[yy]++;
		v[xx].pb(yy);
	}
	
	for(int i=1; i<=n; i++){
		if(!raod[i])root = i;
	}
	
	go(root,0);
	
	for(int i=1; i<=n; i++)cout << p[i] << endl;
	
	
	return 0;
}
```