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;
void getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
priority_queue<int> Q;
int nxt[500010], hed[500010], nedge, too[500010], label[500010], now, n, m, x,
    y, c[500010];
void ae(int x, int y) {
  nxt[++nedge] = hed[x];
  hed[x] = nedge;
  too[nedge] = y;
}
void topsort() {
  for (int i = 1; i <= n; ++i)
    if (c[i] == 0) Q.push(i);
  now = n + 1;
  while (!Q.empty()) {
    int x = Q.top();
    Q.pop();
    label[x] = --now;
    for (int i = hed[x]; i; i = nxt[i]) {
      int y = too[i];
      --c[y];
      if (!c[y]) Q.push(y);
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    getint(x), getint(y);
    ++c[x];
    ae(y, x);
  }
  topsort();
  for (int i = 1; i <= n; ++i) printf("%d ", label[i]);
  return 0;
}
