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;
long long int n, m, cnt = 1;
vector<long long int> G[100005];
vector<long long int> DG[100005];
long long int indegree[100005], label[100005], mark[100005];
priority_queue<long long int> P;
void solve(long long int a) {
  priority_queue<long long int> pq;
  queue<long long int> q;
  q.push(a);
  mark[a] = 1;
  while (!q.empty()) {
    long long int u = q.front();
    q.pop();
    if (indegree[u] == 0) pq.push(u * -1);
    for (int i = 0; i < G[u].size(); i++) {
      long long int v = G[u][i];
      if (mark[v] == 0) {
        q.push(v);
        mark[v] = 1;
      }
    }
  }
  while (!pq.empty()) {
    long long int u = pq.top() * -1;
    pq.pop();
    label[u] = cnt;
    cnt++;
    for (int i = 0; i < DG[u].size(); i++) {
      long long int v = DG[u][i];
      indegree[v]--;
      if (indegree[v] == 0) pq.push(v * -1);
    }
  }
}
void solve2(long long int u) {
  sort(DG[u].begin(), DG[u].end());
  for (int i = 0; i < DG[u].size(); i++) {
    long long int v = DG[u][i];
    if (label[v] == 0) {
      solve2(v);
    }
  }
  label[u] = cnt;
  cnt++;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    long long int a, b;
    cin >> a >> b;
    DG[b].push_back(a);
    indegree[a]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++) {
    if (indegree[i] == 0) P.push(i);
  }
  while (!P.empty()) {
    label[P.top()] = cnt;
    cnt--;
    long long int u = P.top();
    P.pop();
    for (int i = 0; i < DG[u].size(); i++) {
      long long int v = DG[u][i];
      indegree[v]--;
      if (indegree[v] == 0) P.push(v);
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << label[i] << " ";
  }
  cout << endl;
  return 0;
}
