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;
using ll = long long;

const int N = 1e5;
vector<int> g[N];

void dfs2(int i, int p, vector<int> &v) {
  v.emplace_back(g[i].size());
  for(int j : g[i]) if(j != p && g[j].size() >= 2) {
    dfs2(j, i, v);
    return;
  }
}

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n; cin >> n;
  if(n == 2) return (cout << "1 2" << endl, 0);
  for(int i = 0; i < n - 1; i++) {
    int a, b; cin >> a >> b;
    a--; b--;
    g[a].emplace_back(b);
    g[b].emplace_back(a);
  }
  int t = -1;
  for(int i = 0; i < n; i++) {
    if(g[i].size() >= 2) {
      int c = 0;
      for(int j : g[i]) {
        if(g[j].size() >= 2) c++;
      }
      if(c <= 1) t = i;
      if(c >= 3) return (cout << -1 << endl, 0);
    }
  }
  vector<int> v;
  dfs2(t, -1, v);
  int k = v.size();
  for(int i = 0; i < k; i++) {
    if(v[i] == v[k - 1 - i]) continue;
    if(v[i] > v[k - 1 - i]) {
      reverse(begin(v), end(v));
      break;
    }
    break;
  }
  cout << 1;
  int c = 2;
  for(int i = 0; i < k; i++) {
    if(i == k - 1) {
      int t = c++;
      for(int j = 0; j < v[i] - 2; j++) cout << " " << c++;
      cout << " " << t;
      cout << " " << c++;
    } else {
      int t = c++;
      for(int j = 0; j < v[i] - 2; j++) cout << " " << c++;
      cout << " " << t;
    }
  }
  cout << endl;
}

