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;
struct nods {
  int begin, end;
} dot[100005];
int label[100005];
int indegree[100005];
priority_queue<int> s;
vector<int> ve[100005];
void init() {
  memset(indegree, 0, sizeof indegree);
  memset(label, 0, sizeof label);
}
int main() {
  int n, m;
  cin >> n >> m;
  init();
  for (int i = 1; i <= m; i++) {
    cin >> dot[i].end >> dot[i].begin;
    indegree[dot[i].end]++;
    ve[dot[i].begin].push_back(dot[i].end);
  }
  for (int i = 1; i <= n; i++) {
    if (!indegree[i]) s.push(i);
  }
  int t = n;
  while (!s.empty()) {
    int now = s.top();
    s.pop();
    label[now] = t--;
    for (int i = 0; i < ve[now].size(); i++) {
      indegree[ve[now][i]]--;
      if (!indegree[ve[now][i]]) s.push(ve[now][i]);
    }
  }
  for (int i = 1; i <= n; i++) cout << label[i] << endl;
}
