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;
int deg[100001];
vector<int> gra[100001];
int ans[100001];
void topsort() {
  memset(ans, 0, sizeof(ans));
  priority_queue<int, vector<int>, less<int> > q;
  int cnt = n;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 0) {
      q.push(i);
    }
  while (!q.empty()) {
    int h = q.top();
    q.pop();
    ans[h] = cnt--;
    if (gra[h].size() == 0) continue;
    for (int i = 0; i < gra[h].size(); i++) {
      if (--deg[gra[h][i]] == 0) q.push(gra[h][i]);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n >> m) {
    memset(gra, 0, sizeof(gra));
    memset(deg, 0, sizeof(deg));
    for (int i = 1; i <= n; i++) gra[i].clear();
    for (int i = 0; i < m; i++) {
      int a, b;
      cin >> a >> b;
      gra[b].push_back(a);
      deg[a]++;
    }
    topsort();
    for (int i = 1; i <= n; i++)
      if (i != n)
        cout << ans[i] << " ";
      else
        cout << ans[i] << endl;
  }
  return 0;
}
