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 long long maxn = 1e5 + 500;
long long out[maxn];
vector<long long> in[maxn];
vector<long long> topol;
long long ans[maxn];
bool vis[maxn];
set<long long> st;
int main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long v, u;
    cin >> v >> u;
    v--;
    u--;
    out[v]++;
    in[u].push_back(v);
  }
  for (long long i = 0; i < n; i++) {
    if (out[i] == 0) {
      st.insert(-i);
    }
  }
  while (!st.empty()) {
    long long v = (*(st.begin()));
    st.erase(v);
    v = (-v);
    topol.push_back(v);
    for (long long i = 0; i < in[v].size(); i++) {
      long long u = in[v][i];
      out[u]--;
      if (out[u] == 0) {
        st.insert(-u);
      }
    }
  }
  reverse(topol.begin(), topol.end());
  for (long long i = 0; i < n; i++) {
    ans[topol[i]] = i;
  }
  for (long long i = 0; i < n; i++) {
    cout << ans[i] + 1 << ' ';
  }
}
