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 d[100010], a[100010];
vector<int> vec[100010];
priority_queue<int> q;
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    int i, j;
    memset(d, 0, sizeof(d));
    while (m--) {
      int u, v;
      scanf("%d%d", &u, &v);
      d[u]++;
      vec[v].push_back(u);
    }
    for (i = 1; i <= n; i++)
      if (d[i] == 0) q.push(i);
    for (i = n; q.size();) {
      int p = q.top();
      q.pop();
      a[p] = i;
      i--;
      for (j = 0; j < vec[p].size(); j++) {
        int t = vec[p][j];
        d[t]--;
        if (d[t] == 0) q.push(t);
      }
    }
    for (i = 1; i < n; i++) printf("%d ", a[i]);
    printf("%d\n", a[n]);
  }
}
