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;
static const int MAX_N = 1e5 + 5;
int indegree[2 * MAX_N];
int res[2 * MAX_N];
vector<int> V[2 * MAX_N];
priority_queue<int, vector<int>, less<int> > pq;
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(indegree, 0, sizeof(indegree));
    for (int i = 0; i < 2 * MAX_N; i++) {
      V[i].clear();
    }
    int v, u;
    for (int i = 0; i < m; i++) {
      scanf("%d%d", &v, &u);
      V[u].push_back(v);
      indegree[v]++;
    }
    for (int i = 1; i <= n; i++) {
      if (!indegree[i]) {
        pq.push(i);
      }
    }
    int cur = n;
    while (!pq.empty()) {
      u = pq.top();
      pq.pop();
      res[u] = cur--;
      for (int i = 0; i < V[u].size(); i++) {
        v = V[u][i];
        indegree[v]--;
        if (!indegree[v]) {
          pq.push(v);
        }
      }
    }
    for (int i = 1; i < n; i++) {
      printf("%d ", res[i]);
    }
    printf("%d\n", res[n]);
  }
  return 0;
}
