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

typedef long long ll;
typedef vector<int> vi;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

//グラフの隣接リスト
map<int, vector<int>> g;
//頂点の入次数を管理
int h[100010];

int main()
{
  //頂点数v、辺の数e
  int v, e;
  cin >> v >> e;
  REP(i, e) {
    int s, t;
    cin >> s >> t;
    g[s].push_back(t);
    h[t]++;
  }

  //入次数が0の頂点の集合
  stack<int> st;
  REP(i, v) if(h[i] == 0) st.push(i);

  vi ans;
  
  while(st.size()) { //stがなくなるまでループ
    //stの集合のから一つ取り出す
    int i = st.top(); st.pop();
    ans.push_back(i);
    for(auto& j: g[i]) {
      h[j]--; //隣接する頂点の入次数をマイナス1
      //これによって入次数が0になればstに追加
      if(h[j] == 0) st.push(j);
    }
  }

  for(int i: ans) cout << i << endl; //ans.size() == v であればトポロジカルソートができた -> 閉路なし
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

<image>


A directed acyclic graph (DAG) can be used to represent the ordering of tasks. Tasks are represented by vertices and constraints where one task can begin before another, are represented by edges. For example, in the above example, you can undertake task B after both task A and task B are finished. You can obtain the proper sequence of all the tasks by a topological sort.

Given a DAG $G$, print the order of vertices after the topological sort.

Constraints

* $1 \leq |V| \leq 10,000$
* $0 \leq |E| \leq 100,000$
* There are no parallel edges in $G$
* There are no self loops in $G$

Input

A directed graph $G$ is given in the following format:
$|V|\;|E|$
$s_0 \; t_0$
$s_1 \; t_1$
:
$s_{|E|-1} \; t_{|E|-1}$



$|V|$ is the number of vertices and $|E|$ is the number of edges in the graph. The graph vertices are named with the numbers $0, 1,..., |V|-1$ respectively.

$s_i$ and $t_i$ represent source and target nodes of $i$-th edge (directed).

Output

Print the vertices numbers in order. Print a number in a line.

If there are multiple possible solutions, print any one of them (the solution is judged by a special validator).

Example

Input

6 6
0 1
1 2
3 1
3 4
4 5
5 2


Output

0
3
1
4
5
2

### Response

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

typedef long long ll;
typedef vector<int> vi;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

//グラフの隣接リスト
map<int, vector<int>> g;
//頂点の入次数を管理
int h[100010];

int main()
{
  //頂点数v、辺の数e
  int v, e;
  cin >> v >> e;
  REP(i, e) {
    int s, t;
    cin >> s >> t;
    g[s].push_back(t);
    h[t]++;
  }

  //入次数が0の頂点の集合
  stack<int> st;
  REP(i, v) if(h[i] == 0) st.push(i);

  vi ans;
  
  while(st.size()) { //stがなくなるまでループ
    //stの集合のから一つ取り出す
    int i = st.top(); st.pop();
    ans.push_back(i);
    for(auto& j: g[i]) {
      h[j]--; //隣接する頂点の入次数をマイナス1
      //これによって入次数が0になればstに追加
      if(h[j] == 0) st.push(j);
    }
  }

  for(int i: ans) cout << i << endl; //ans.size() == v であればトポロジカルソートができた -> 閉路なし
  return 0;
}

```