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 M = 1e5 + 44;
int n, m;
int cnt[M];
int label[M];
vector<int> v[M];
priority_queue<int, vector<int>, less<int> > q;
int main() {
  int i, j, a, b, tmp;
  cin >> n >> m;
  for (i = 1; i <= n; i++) v[i].clear();
  memset(cnt, 0, sizeof(cnt));
  memset(label, -1, sizeof(label));
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    v[b].push_back(a);
    cnt[a]++;
  }
  while (!q.empty()) q.pop();
  for (i = 1; i <= n; i++)
    if (cnt[i] == 0) q.push(i);
  int now = n, p, sz;
  while (!q.empty()) {
    p = q.top();
    q.pop();
    label[p] = now;
    now--;
    sz = v[p].size();
    for (i = 0; i < sz; i++) {
      tmp = v[p][i];
      cnt[tmp]--;
      if (cnt[tmp] == 0) q.push((tmp));
    }
  }
  for (i = 1; i <= n; i++) {
    if (i - 1) printf(" ");
    printf("%d", label[i]);
  }
  printf("\n");
  return 0;
}
