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;
struct node {
  int d, number;
  friend bool operator<(struct node a, struct node b) {
    return a.number < b.number;
  }
} v[100005];
vector<int> e[100005];
int label[100005];
priority_queue<struct node> q;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    v[i].number = i;
  }
  for (int i = 0; i < m; i++) {
    int sta, en;
    scanf("%d%d", &sta, &en);
    v[sta].d++;
    e[en].push_back(sta);
  }
  for (int i = 1; i <= n; i++) {
    if (v[i].d == 0) {
      q.push(v[i]);
    }
  }
  int cnt = n;
  while (!q.empty()) {
    struct node temp = q.top();
    q.pop();
    label[temp.number] = cnt--;
    for (int i = 0; i < e[temp.number].size(); i++) {
      int x = e[temp.number][i];
      if (v[x].d) {
        v[x].d--;
        if (v[x].d == 0) q.push(v[x]);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << label[i] << " ";
  }
  return 0;
}
