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;
const int MOD = 1e9 + 7;
const int INF = 0x7ffffff;
const int magic = 348;
priority_queue<int> q;
vector<int> v[100048];
int cnt[100048], ans[100048];
int n, e;
int main() {
  int i, first, second;
  scanf("%d%d", &n, &e);
  for (i = 1; i <= e; i++) {
    scanf("%d%d", &first, &second);
    v[second].push_back(first);
    cnt[first]++;
  }
  for (i = 1; i <= n; i++)
    if (cnt[i] == 0) q.push(i);
  int tot = n;
  while (!q.empty()) {
    first = q.top();
    q.pop();
    ans[first] = tot--;
    for (i = 0; i < v[first].size(); i++) {
      cnt[v[first][i]]--;
      if (cnt[v[first][i]] == 0) q.push(v[first][i]);
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
