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 Size = 1024;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 2e6 + 15;
int head[maxn];
int ver[maxn];
int nxt[maxn];
int tot;
void add(int x, int y) {
  ver[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
int in[maxn];
int num[maxn];
int main() {
  int n, m, a, b;
  scanf("%d%d", &n, &m);
  for (int i = 1; m >= i; i++) {
    scanf("%d%d", &a, &b);
    add(b, a);
    in[a]++;
  }
  priority_queue<int> q;
  for (int i = 1; n >= i; i++) {
    if (!in[i]) q.push(i);
  }
  int cnt = n;
  while (q.size()) {
    int x = q.top();
    q.pop();
    num[x] = cnt--;
    for (int i = head[x]; i; i = nxt[i]) {
      int y = ver[i];
      in[y]--;
      if (!in[y]) q.push(y);
    }
  }
  int tt = 0;
  for (int i = 1; n >= i; i++) {
    if (tt++ > 0) printf(" ");
    printf("%d", num[i]);
  }
  printf("\n");
  return 0;
}
