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 + 9;
int head[maxn], Next[maxn], ver[maxn], deg[maxn], k, cnt, a[maxn];
int n, m;
void topsort() {
  priority_queue<int> q;
  priority_queue<int> Q;
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 0) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int x = q.top();
    q.pop();
    a[x] = cnt--;
    for (int i = head[x]; i; i = Next[i]) {
      int y = ver[i];
      if ((--deg[y]) == 0) q.push(y);
    }
  }
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    k = 0;
    cnt = n;
    for (int i = 1; i <= m; i++) {
      int y, x;
      scanf("%d%d", &y, &x);
      ver[i] = y;
      Next[i] = head[x], head[x] = i;
      deg[y]++;
    }
    topsort();
    for (int i = 1; i <= n; i++) {
      if (i != n)
        printf("%d ", a[i]);
      else
        printf("%d\n", a[i]);
    }
  }
  return 0;
}
