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 = 1e5 + 10;
struct edge {
  int v;
  int Next;
} Edge[maxn * 10];
int tot = 0;
int head[maxn];
int ind[maxn];
int ans[maxn];
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void addedge(int u, int v) {
  Edge[tot].v = v;
  Edge[tot].Next = head[u];
  head[u] = tot++;
}
int n, m;
int main() {
  cin >> n >> m;
  init();
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    addedge(v, u);
    ind[u]++;
  }
  priority_queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (ind[i] == 0) q.push(i);
  }
  int cnt = n;
  while (!q.empty()) {
    int x = q.top();
    q.pop();
    ans[x] = cnt--;
    for (int i = head[x]; i != -1; i = Edge[i].Next) {
      int v = Edge[i].v;
      ind[v]--;
      if (ind[v] == 0) {
        q.push(v);
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
