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 = 1e5 + 111;
int n, m;
int deg[N], res[N];
struct cmp {
  bool operator()(int x, int y) {
    if (deg[x] == deg[y]) return x > y;
    return deg[x] < deg[y];
  }
};
set<int, cmp> St;
set<int, cmp>::iterator it;
vector<int> Adj[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  while (m--) {
    int u, v;
    cin >> u >> v;
    Adj[v].push_back(u);
    deg[u]++;
  }
  for (int i = 1; i <= n; i++) {
    St.insert(i);
  }
  int lab = n;
  while (!St.empty()) {
    int u = *St.begin();
    St.erase(St.begin());
    for (int i = 0; i < Adj[u].size(); i++) {
      int v = Adj[u][i];
      it = St.find(v);
      St.erase(it);
      deg[v]--;
      St.insert(v);
    }
    res[u] = lab;
    lab--;
  }
  for (int i = 1; i <= n; i++) cout << res[i] << " ";
  return 0;
}
