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;
vector<int> G[100010];
struct P {
  int x;
  bool operator<(const P &hs) const { return x > hs.x; }
};
priority_queue<int> Q;
int d[100010], ans[100010];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  memset(d, 0, sizeof(d));
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    G[y].push_back(x);
    d[x]++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i] == 0) {
      Q.push(i);
    }
  int cn = n;
  while (!Q.empty()) {
    int x = Q.top();
    Q.pop();
    ans[x] = cn--;
    for (int i = 0; i < G[x].size(); i++) {
      int v = G[x][i];
      d[v]--;
      if (d[v] == 0) Q.push(v);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
