#include<bits/stdc++.h>
using namespace std;

typedef pair<int,int> pii;

priority_queue<pii> pq;

vector<int> adj[100005];
int n,m;
int par[100005];
int deg[100005];

int main(){
   cin >> n >> m;
   for(int i=1;i<n+m;i++){
      int u,v;
      cin >> u >> v;
      deg[v]++;
      adj[u].push_back(v);
   }
   for(int i=1;i<=n;i++){
      pii tmp;
      tmp.first = -deg[i];
      tmp.second = i;
      pq.push(tmp);
   }
   for(int i=1;i<=n;i++){
      pii lol = pq.top(); pq.pop();
      int cur = lol.second;
      for(int j=0;j<adj[cur].size();j++){
         int now = adj[cur][j];
         if(deg[now] == 1){
            par[now] = cur;
            pq.push({0,now});
         } else {
            deg[now]--;
         }
      }
   }
   for(int i=1;i<=n;i++) cout << par[i] << endl;
	return 0;
}

### Prompt

Please create a solution in CPP 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> pii;

priority_queue<pii> pq;

vector<int> adj[100005];
int n,m;
int par[100005];
int deg[100005];

int main(){
   cin >> n >> m;
   for(int i=1;i<n+m;i++){
      int u,v;
      cin >> u >> v;
      deg[v]++;
      adj[u].push_back(v);
   }
   for(int i=1;i<=n;i++){
      pii tmp;
      tmp.first = -deg[i];
      tmp.second = i;
      pq.push(tmp);
   }
   for(int i=1;i<=n;i++){
      pii lol = pq.top(); pq.pop();
      int cur = lol.second;
      for(int j=0;j<adj[cur].size();j++){
         int now = adj[cur][j];
         if(deg[now] == 1){
            par[now] = cur;
            pq.push({0,now});
         } else {
            deg[now]--;
         }
      }
   }
   for(int i=1;i<=n;i++) cout << par[i] << endl;
	return 0;
}

```