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;
int n, m, d[222222], tot, ans[222222];
int TOT, EN[222222], NE[222222], LA[222222];
void ADD(int x, int y) {
  TOT++;
  EN[TOT] = y;
  NE[TOT] = LA[x];
  LA[x] = TOT;
}
priority_queue<int> q;
int main() {
  int i, j, x, y;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    d[x]++;
    ADD(y, x);
  }
  for (i = 1; i <= n; i++)
    if (d[i] == 0) q.push(i);
  tot = n;
  while (!q.empty()) {
    x = q.top();
    q.pop();
    ans[x] = tot--;
    for (i = LA[x]; i; i = NE[i]) {
      y = EN[i];
      if (d[y] > 0) {
        d[y]--;
        if (d[y] == 0) q.push(y);
      }
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", ans[i]);
}
