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 read() {
  int w = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9') c = getchar(), f = c == '-' ? -1 : f;
  while (c >= '0' && c <= '9') w = w * 10 + c - 48, c = getchar();
  return w * f;
}
int n, m, in[100005], ans[100005], minn[100005], p[100005], cnt;
bool vis[100005];
vector<int> e[100005], w[100005], E[100005];
void dfs(int u) {
  w[cnt].push_back(u);
  vis[u] = 1;
  minn[cnt] = min(minn[cnt], u);
  for (auto v : E[u])
    if (!vis[v]) dfs(v);
}
signed main() {
  n = read(), m = read();
  for (int i = 1; i <= m; i++) {
    int x = read(), y = read();
    E[x].push_back(y);
    E[y].push_back(x);
    e[y].push_back(x);
    in[x]++;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) minn[i] = 100005, cnt++, dfs(i);
  for (int i = 1; i <= cnt; i++) p[i] = i;
  int now = n;
  priority_queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i);
  while (!q.empty()) {
    int u = q.top();
    q.pop();
    ans[u] = now--;
    for (auto v : e[u]) {
      in[v]--;
      if (!in[v]) q.push(v);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
