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> > adj;
vector<int> ind;
priority_queue<int> pq;
vector<int> ans;
int main() {
  scanf("%d %d", &N, &M);
  adj.resize(N);
  ind = vector<int>(N, 0);
  for (int i = 0; i < M; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    adj[v].push_back(u);
    ind[u]++;
  }
  for (int i = 0; i < N; i++)
    if (!ind[i]) pq.push(i);
  ans.resize(N);
  int num = N;
  while (num) {
    int u = pq.top();
    pq.pop();
    ans[u] = num--;
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      ind[v]--;
      if (!ind[v]) pq.push(v);
    }
  }
  for (int i = 0; i < N; i++) {
    printf("%d ", ans[i]);
  }
}
