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 maxn = 100007;
int n, m, du[maxn], a[maxn], cnt;
vector<int> G[maxn];
bool inq[maxn];
priority_queue<int> q;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    static int u, v;
    scanf("%d%d", &u, &v);
    G[v].push_back(u);
    du[u]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++) {
    if (du[i] == 0) {
      inq[i] = true;
      q.push(i);
    }
  }
  while (!q.empty()) {
    static int p;
    p = q.top();
    q.pop();
    a[p] = cnt--;
    for (int i = 0; i < G[p].size(); i++) {
      static int t;
      t = G[p][i];
      if (!inq[t]) {
        du[t]--;
        if (du[t] == 0) {
          q.push(t);
          inq[t] = true;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  puts("");
  return 0;
}
