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 = 2500007;
int head[maxn], sum;
struct EDGE {
  int to;
  int next;
  int from;
} edge[maxn];
void add(int x, int y) {
  edge[++sum].next = head[x];
  edge[sum].from = x;
  edge[sum].to = y;
  head[x] = sum;
}
int n, m, x, y, tot;
int in[maxn], ans[maxn];
void topu() {
  priority_queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (in[i] == 0) q.push(i);
  while (!q.empty()) {
    int k = q.top();
    q.pop();
    ans[k] = tot--;
    for (int i = head[k]; i; i = edge[i].next) {
      int v = edge[i].to;
      in[v]--;
      if (in[v] == 0) q.push(v);
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &x, &y);
    add(y, x);
    in[x]++;
  }
  tot = n;
  topu();
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
}
