You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.

You should assign labels to all vertices in such a way that:

  * Labels form a valid permutation of length n — an integer sequence such that each integer from 1 to n appears exactly once in it. 
  * If there exists an edge from vertex v to vertex u then labelv should be smaller than labelu. 
  * Permutation should be lexicographically smallest among all suitable. 



Find such sequence of labels to satisfy all the conditions.

Input

The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).

Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges.

Output

Print n numbers — lexicographically smallest correct permutation of labels of vertices.

Examples

Input

3 3
1 2
1 3
3 2


Output

1 3 2 


Input

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


Output

4 1 2 3 


Input

5 4
3 1
2 1
2 3
4 5


Output

3 1 2 4 5 
#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<vector<int> > grafo;
vector<int> indegre;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  grafo.resize(n, vector<int>());
  indegre.resize(n, 0);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    grafo[v].push_back(u);
    indegre[u]++;
  }
  priority_queue<int> S;
  for (int i = 0; i < n; i++) {
    if (indegre[i] == 0) {
      S.push(i);
    }
  }
  vector<int> L;
  L.resize(n, 0);
  int pos = n - 1;
  while (S.size() != 0) {
    int v = S.top();
    S.pop();
    L[v] = pos;
    for (int w = 0; w < grafo[v].size(); w++) {
      int vecino = grafo[v][w];
      indegre[vecino]--;
      if (indegre[vecino] == 0) {
        S.push(vecino);
      }
    }
    pos--;
  }
  for (int i = 0; i < n; i++) {
    cout << L[i] + 1 << " ";
  }
  cout << "\n";
}
