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
//Link : https://atcoder.jp/contests/arc095/tasks/arc095_d

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100005

vector<int> adj[N];

int numDegTwoNode[N];
bool dfs(int node,int par) {
  numDegTwoNode[node]= 0;
  for(int ch : adj[node]) {
    if(adj[ch].size()>=2) {
      ++numDegTwoNode[node];
    }
    if(ch!=par) {
      if(!dfs(ch,node)) {
        return false;
      }
    }
  }
  if(numDegTwoNode[node]>2) {
    return false;
  }
  return true;
}

int path[N];
int pos = 1;
void dfs2(int node,int par) {
  path[pos++] = node;
  for(int ch : adj[node]) {
    if(ch==par) {
      continue;
    }
    if(adj[ch].size()>=2) {
      dfs2(ch,node);
      break;
    }
  }
}

int per[N];
int tmp[N];

void isBetter(int n) {
  for(int i=0;i<n;++i) {
    if(per[i]!=tmp[i]) {
      if(per[i]>tmp[i]) {
        for(int j = 0;j<n;++j) {
          per[j] = tmp[j];
        }
      }
      return;
    }
  }
}

int deg[N];


void makePermuetation(int size,int n) {
  int index = 1;
  int per = 2;
  tmp[0] = 1;
  for(int i=1;i<size;++i) {
    int sz = deg[i] - 2;
    tmp[index+sz] = per++;
    for(int j =0;j<sz;++j) {
      tmp[index+j] = per++;
    }
    index += sz+1;
  }
  tmp[index] = per;
  assert(per==n);
}
void solve() {
  int n;scanf("%d ", &n);
  if(n==2) {
    printf("1 2\n");
    return;
  }
  for(int i=0;i<n-1;++i) {
    int a,b;scanf("%d %d ", &a,&b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  if(!dfs(1,0)) {
    printf("-1\n");
    return;
  }

  int stNode = -1;
  for(int i=1;i<=n;++i) {
    if(adj[i].size()>=2 && numDegTwoNode[i]<=1) {
      stNode = i;
      break;
    }
  }
  assert(stNode!=-1);
  dfs2(stNode,0);
  for(int i=1;i<pos;++i) {
    deg[i] = adj[path[i]].size();
  }
  per[0] = n+2;
  makePermuetation(pos,n);
  isBetter(n);
  reverse(deg+1,deg+pos);
  makePermuetation(pos,n);
  isBetter(n);

  for(int i=0;i<n;++i) {
    printf("%d ", per[i]);
  }
}

int main() {
  //freopen("input.txt","r",stdin);
  solve();
  return 0;
}
