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 main() {
  int n, m;
  scanf("%d %d", &n, &m);
  vector<vector<int>> g(n);
  vector<vector<int>> rg(n);
  vector<int> cnt(n);
  for (int i = 0; i < m; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    --x;
    --y;
    g[x].push_back(y);
    rg[y].push_back(x);
    ++cnt[x];
  }
  set<int> q;
  for (int i = 0; i < n; ++i) {
    if (cnt[i] == 0) q.insert(i);
  }
  vector<int> ans(n);
  int cur = n;
  while (!q.empty()) {
    int v = *q.rbegin();
    ans[v] = cur--;
    q.erase(--q.end());
    for (int to : rg[v]) {
      --cnt[to];
      if (cnt[to] == 0) q.insert(to);
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", ans[i]);
  return 0;
}
