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;
const int MAXM = 1e5 + 10;
struct Edge {
  int to, nxt;
};
int n, m, cnt_e;
int head[MAXN];
int deg[MAXN];
Edge e[MAXM << 1];
inline void addedge(int u, int v) {
  e[++cnt_e].to = v;
  e[cnt_e].nxt = head[u];
  head[u] = cnt_e;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(v, u);
    deg[u]++;
  }
  static int ans[MAXN];
  priority_queue<int> pq;
  for (int i = 1; i <= n; i++)
    if (!deg[i]) pq.push(i);
  int cnt = n;
  while (!pq.empty()) {
    int u = pq.top();
    pq.pop();
    ans[u] = cnt--;
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      deg[v]--;
      if (!deg[v]) pq.push(v);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
