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;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const ll MOD = 1000000007;
const ll HIGHINF = (ll)1e18;
const int INF = 1e9;
int n, m;
V<V<int> > edges;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> m;
  edges.resize(n);
  V<int> h(n, 0);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    edges[v].emplace_back(u);
    h[u]++;
  }
  set<int> st;
  for (int i = 0; i < n; i++) {
    if (h[i] == 0) st.insert(i);
  }
  V<int> tps;
  while (!st.empty()) {
    int idx = *prev(st.end());
    st.erase(prev(st.end()));
    tps.emplace_back(idx);
    for (int e : edges[idx]) {
      h[e]--;
      if (h[e] == 0) st.insert(e);
    }
  }
  V<int> ans(n);
  for (int i = 0; i < n; i++) ans[tps[i]] = n - i;
  for (int i = 0; i < n; i++) cout << ans[i] << (i == n - 1 ? '\n' : ' ');
  return 0;
}
