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 N = 1e5 + 5;
vector<int> g[N];
priority_queue<int> q;
int in[N], label[N];
int who;
void kahn() {
  while (!q.empty()) {
    int u = q.top();
    q.pop();
    label[u] = who--;
    for (int &to : g[u]) {
      --in[to];
      if (in[to] == 0) {
        q.emplace(to);
      }
    }
  }
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[y].emplace_back(x);
    ++in[x];
  }
  for (int i = 1; i <= n; ++i) {
    if (!in[i]) {
      q.emplace(i);
    }
  }
  who = n;
  kahn();
  for (int i = 1; i <= n; ++i) {
    printf("%d ", label[i]);
  }
  return 0;
}
