Takahashi has an ability to generate a tree using a permutation (p_1,p_2,...,p_n) of (1,2,...,n), in the following process:

First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n, perform the following operation:

* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.



Takahashi is trying to make his favorite tree with this ability. His favorite tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite tree by using a proper permutation. If he can do so, find the lexicographically smallest such permutation.

Constraints

* 2 \leq n \leq 10^5
* 1 \leq v_i, w_i \leq n
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1}


Output

If there is no permutation that can generate a tree isomorphic to Takahashi's favorite tree, print `-1`. If it exists, print the lexicographically smallest such permutation, with spaces in between.

Examples

Input

6
1 2
1 3
1 4
1 5
5 6


Output

1 2 4 5 3 6


Input

6
1 2
2 3
3 4
1 5
5 6


Output

1 2 3 4 5 6


Input

15
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
5 10
5 11
6 12
6 13
7 14
7 15


Output

-1
#include <bits/stdc++.h>
using namespace std;

int dep[101000], fa[101000], root, rfa[101000], ans[101000], cnt[101000];
int pnt, in[101000], x, y, n;
vector<int> g[101000];

void Dfs(int x) {
  //cerr << x << endl;
  if(dep[root] < dep[x])
    root = x;
  for(int i = 0; i < g[x].size(); i++) {
    int v = g[x][i];
    if(v == fa[x]) continue;
    dep[v] = dep[x] + 1;
    fa[v] = x;
    Dfs(v);
  }
}

void Out(int x, int *a) {
  int now = 1, pnt = 0;
  for(; x; x = a[x]) {
    for(int i = now + 1; i <= now + cnt[x]; i++)
      ans[i - 1] = i;
    ans[now + cnt[x]] = now; 
    now += cnt[x] + 1;
  }
}

bool Solve(void) {
  if(dep[root] == 1) {
    ans[1] = 1, ans[2] = 2;
    return 1;
  }
  int head = root, tail = root;
  while(root) {
    in[root] = 1;
    tail = root;
    if(fa[root]) rfa[fa[root]] = root;
    root = fa[root];
  }
  cerr << head << " " << tail << endl;
  for(int i = 1; i <= n; i++)
    if(!in[i] && !in[fa[i]]) return 0;
    else if(!in[i]) ++cnt[fa[i]];
  bool flg = 0;
  for(int h = head, t = tail; h != tail; h = fa[h], t = rfa[t])
    if(cnt[h] != cnt[t]) {
      if(cnt[h] < cnt[t])
	Out(head, fa);
      else Out(tail, rfa);
      flg = 1;
      break;
    }
  if(!flg) Out(head, fa);
  return 1;
}

int main(void) {
  scanf("%d", &n);
  for(int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  Dfs(1);
  //cerr << root << endl;
  fa[root] = 0;
  dep[root] = 0;
  Dfs(root);
  //cerr << root << endl;
  if(Solve())
    for(int i = 1; i <= n; i++)
      printf("%d ", ans[i]);
  else puts("-1");
  return 0;
}
