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, x, y, layer[100001], cnt, visit[100001];
priority_queue<int> p;
vector<int> v[100001];
void bfs() {
  while (!p.empty()) {
    int u = p.top();
    layer[u] = cnt;
    cnt--;
    p.pop();
    for (int i = 0; i < v[u].size(); ++i) {
      int uu = v[u][i];
      if (layer[uu] != 0) continue;
      visit[uu]--;
      if (visit[uu] == 0) p.push(uu);
    }
  }
}
int main() {
  cin >> n >> m;
  cnt = n;
  while (m--) {
    cin >> x >> y;
    v[y].push_back(x);
    visit[x]++;
  }
  for (int i = 1; i <= n; ++i)
    if (visit[i] == 0) p.push(i);
  bfs();
  for (int i = 1; i <= n; ++i) cout << layer[i] << ' ';
}
