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 long long MOD = 1e9 + 7;
const long long INF = 1LL << 32;
const double eps = 1e-13;
int d[100009];
vector<int> adj[100009];
int p[100009];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (m); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    adj[b].push_back(a);
    d[a]++;
  }
  priority_queue<int> pq;
  for (int i = 0; i < (n); ++i) {
    if (d[i] == 0) pq.push(i);
  }
  int ct = n;
  while (((int)(pq).size())) {
    int x = pq.top();
    pq.pop();
    p[x] = ct--;
    for (int i : adj[x]) {
      d[i]--;
      if (d[i] == 0) pq.push(i);
    }
  }
  for (int i = 0; i < (n); ++i) printf("%d ", p[i]);
}
