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>
const int N = 1 << 18;
const int M = 1 << 18;
int n, m;
std::set<int> g[N];
std::vector<int> pg[N];
int lbl[N];
void myassert(bool b) {
  if (!b) printf("%d\n", 1 / b);
}
int main(int argc, char* argv[]) {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].insert(v);
    pg[v].push_back(u);
  }
  std::set<int, std::greater<int> > ss;
  for (int i = 1; i <= n; ++i) {
    if (g[i].empty()) {
      ss.insert(i);
    }
  }
  int cur_label = n;
  while (!ss.empty()) {
    int u = *ss.begin();
    lbl[u] = cur_label--;
    ss.erase(ss.begin());
    for (int p : pg[u]) {
      g[p].erase(u);
      if (g[p].empty()) {
        ss.insert(p);
      }
    }
  }
  myassert(cur_label == 0);
  std::vector<int> visit(1 + n);
  for (int i = 1; i <= n; ++i) {
    myassert(lbl[i] != 0);
    visit[lbl[i]] = 1;
  }
  for (int i = 1; i <= n; ++i) myassert(visit[i] == 1);
  for (int i = 1; i <= n; ++i) printf("%d ", lbl[i]);
  return 0;
}
