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 M = 500000;
int n, m, t;
int a[M], ans[M];
struct edge {
  int u, next, last;
} e[M];
void add(int u, int v) {
  t++;
  e[t].u = v;
  e[t].next = e[u].last;
  e[u].last = t;
}
int main() {
  int x, y;
  t = 0;
  priority_queue<int> q;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    a[x]++;
    add(y, x);
  }
  for (int i = 1; i <= n; i++)
    if (a[i] == 0) q.push(i);
  int j = n;
  while (q.size()) {
    x = q.top();
    q.pop();
    ans[x] = j--;
    for (int i = e[x].last; i; i = e[i].next) {
      y = e[i].u;
      if (a[y] > 0) a[y]--;
      if (a[y] == 0) q.push(y);
    }
  }
  for (int i = 1; i < n; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[n]);
  return 0;
}
