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 int maxn = 100010;
const int maxm = 100010;
struct Arc {
  int to, next;
} arcs[maxm];
int head[maxn], ne, out[maxn], ans[maxn];
inline void add_arc(int u, int v);
priority_queue<int> Q;
int main() {
  int n, m, i, u, v, k;
  while (scanf("%d%d", &n, &m) > 0) {
    memset(head, -1, sizeof(int) * (n + 1));
    ne = 0;
    memset(out, 0, sizeof(int) * (n + 1));
    for (i = 0; i < m; i++) {
      scanf("%d%d", &u, &v);
      add_arc(v, u);
      out[u]++;
    }
    while (!Q.empty()) Q.pop();
    for (i = 1; i <= n; i++) {
      if (out[i] == 0) Q.push(i);
    }
    for (k = n; k >= 1; k--) {
      u = Q.top();
      Q.pop();
      ans[u] = k;
      for (i = head[u]; i != -1; i = arcs[i].next) {
        v = arcs[i].to;
        if (--out[v] == 0) Q.push(v);
      }
    }
    for (i = 1; i <= n; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}
inline void add_arc(int u, int v) {
  arcs[ne].to = v;
  arcs[ne].next = head[u];
  head[u] = ne++;
  return;
}
