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 int N = 100000, M = 100000;
vector<int> rev[N + 1];
int deg[N + 1];
int label[N + 1];
priority_queue<int> que;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  memset(deg, 0, sizeof deg);
  for (int f, t, i = 0; i < m; ++i) {
    scanf("%d%d", &f, &t);
    rev[t].push_back(f);
    ++deg[f];
  }
  for (int i = 1; i <= n; ++i)
    if (!deg[i]) que.push(i);
  for (int lab = n, t; lab > 0; --lab) {
    t = que.top();
    que.pop();
    label[t] = lab;
    for (int i = 0; i < rev[t].size(); ++i)
      if (--deg[rev[t][i]] == 0) que.push(rev[t][i]);
  }
  for (int i = 1; i <= n; ++i) printf("%d%c", label[i], i == n ? '\n' : ' ');
  return 0;
}
