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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
const int MAXM = 100000;
int n, m;
int ghead[MAXN], gnxt[MAXM], gto[MAXM];
int rhead[MAXN], rnxt[MAXM], rto[MAXM];
int indeg[MAXN], outdeg[MAXN];
bool need[MAXN];
int ret[MAXN];
priority_queue<int> pq;
void run() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); ++i)
    ghead[i] = rhead[i] = -1, indeg[i] = outdeg[i] = 0, need[i] = false;
  for (int i = (0); i < (m); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    gnxt[i] = ghead[a], ghead[a] = i, gto[i] = b, ++outdeg[a];
    rnxt[i] = rhead[b], rhead[b] = i, rto[i] = a, ++indeg[b];
  }
  memset(ret, -1, sizeof(ret));
  for (int i = (0); i < (n); ++i)
    if (outdeg[i] == 0) pq.push(i);
  int now = n;
  while (((int)(pq).size()) > 0) {
    int at = pq.top();
    pq.pop();
    ret[at] = now--;
    for (int x = rhead[at]; x != -1; x = rnxt[x]) {
      int to = rto[x];
      if (--outdeg[to] == 0) pq.push(to);
    }
  }
  for (int i = (0); i < (n); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ret[i]);
  }
  puts("");
}
int main() {
  run();
  return 0;
}
