#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int n, m, a, b, deg[100010], ans[100010];
vector<int> g[100010];

int main() {
  cin.tie(0); ios_base::sync_with_stdio(false);

  cin >> n >> m;
  for(int i = 0; i < n - 1 + m; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    ++deg[b];
  }

  int root;
  for(int i = 1; i <= n; ++i) if(deg[i] == 0) root = i;

  queue<int> q;
  q.push(root);
  while(q.size()) {
    int v = q.front(); q.pop();
    
    for(int nv : g[v]) {
      --deg[nv];
      if(deg[nv] == 0) {
        q.push(nv);
        ans[nv] = v;
      }
    }
  }
  for(int i = 1; i <= n; ++i) cout << ans[i] << '\n';
  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 <iostream>
#include <vector>
#include <queue>
using namespace std;

int n, m, a, b, deg[100010], ans[100010];
vector<int> g[100010];

int main() {
  cin.tie(0); ios_base::sync_with_stdio(false);

  cin >> n >> m;
  for(int i = 0; i < n - 1 + m; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    ++deg[b];
  }

  int root;
  for(int i = 1; i <= n; ++i) if(deg[i] == 0) root = i;

  queue<int> q;
  q.push(root);
  while(q.size()) {
    int v = q.front(); q.pop();
    
    for(int nv : g[v]) {
      --deg[nv];
      if(deg[nv] == 0) {
        q.push(nv);
        ans[nv] = v;
      }
    }
  }
  for(int i = 1; i <= n; ++i) cout << ans[i] << '\n';
  return 0;
}
```