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 head[100010], cnt, a[100010], vis[100010], ans[100010], n, m, temp;
inline int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - 48, c = getchar();
  return x;
}
struct node {
  int u, v, next;
} edge[100010];
void add(int a, int b) {
  edge[cnt].u = a;
  edge[cnt].v = b;
  edge[cnt].next = head[a];
  head[a] = cnt++;
}
priority_queue<int> q;
int main() {
  memset(head, 0x3f3f3f3f, sizeof(head));
  n = read();
  m = read();
  for (int i = 1; i <= m; i++) {
    int u, v;
    u = read();
    v = read();
    add(v, u);
    a[u]++;
  }
  for (int i = 1; i <= n; i++)
    if (!a[i]) q.push(i);
  temp = n;
  while (!q.empty()) {
    int u = q.top();
    ans[u] = temp;
    temp--;
    vis[u] = true;
    q.pop();
    for (int i = head[u]; i != 0x3f3f3f3f; i = edge[i].next) {
      int nt = edge[i].v;
      if (vis[nt] == true) continue;
      a[nt]--;
      if (a[nt] == 0) q.push(nt);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
