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;
struct A {
  long long to;
  A* nxt;
} a[(20ll + (long long)1e5) * 2];
long long na;
A* head[(20ll + (long long)1e5)];
void ins(long long from, long long to) {
  a[na] = A{to, head[from]};
  head[from] = &a[na++];
}
long long od[(20ll + (long long)1e5)];
long long ass[(20ll + (long long)1e5)];
using pii = pair<long long, long long>;
int main() {
  long long n, m;
  scanf(
      "%lld"
      "%lld",
      &n, &m);
  for (long long i = 0; i < m; i++) {
    long long u, v;
    scanf(
        "%lld"
        "%lld",
        &u, &v);
    ins(v, u);
    od[u]++;
  }
  priority_queue<long long, vector<long long>, less<long long>> q;
  for (long long i = 1; i <= n; i++) {
    if (!od[i]) q.push(i);
  }
  long long idx = n;
  while (!q.empty()) {
    long long v = q.top();
    q.pop();
    if (!ass[v]) ass[v] = idx--;
    A* p = head[v];
    while (p) {
      od[p->to]--;
      if (!od[p->to]) q.push(p->to);
      p = p->nxt;
    }
  }
  for (long long __i = 0; __i < (n); __i++)
    printf(
        "%lld"
        "%c",
        (ass + 1)[__i], __i + 1 == (n) ? 10 : 32);
  ;
  return 0;
}
