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[101111];
int visited[101111];
int label[101111];
int n;
int outdegree[101111];
int main() {
  int i, m;
  scanf("%d%d", &n, &m);
  priority_queue<int> q;
  for (i = 0; i < m; i++) {
    int v1, v2;
    scanf("%d%d", &v1, &v2);
    g[v2].push_back(v1);
    outdegree[v1]++;
  }
  for (i = 1; i <= n; i++) {
    if (!outdegree[i]) {
      q.push(i);
    }
  }
  int ans = n;
  while (!q.empty()) {
    int v = q.top();
    q.pop();
    if (visited[v]) {
      continue;
    }
    label[v] = ans--;
    visited[v] = 1;
    for (auto nv : g[v]) {
      if (!visited[nv]) {
        outdegree[nv]--;
        if (!outdegree[nv]) {
          q.push(nv);
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    printf("%d ", label[i]);
  }
  return 0;
}
