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

int N;
vector<int> A[100000];
int S;
vector<int> L;

bool see(int v, int par) {
  int count = 0;
  for(int w : A[v]) {
    if(w != par && (int)A[w].size() != 1) {
      ++count;
      if(!see(w, v)) return false;
    }
  }
  if(par >= 0 && count > 1) return false;
  else if(par < 0 && count > 2) return false;
  else {
    if(count == 0) S = v;
    return true;
  }
}

void go(int v, int par) {
  L.push_back(A[v].size() - 2);
  for(int w : A[v]) {
    if(w != par && (int)A[w].size() != 1) { go(w, v); return; }
  }
}

void make(const vector<int>& l, vector<int>& r) {
  int c = 0;
  r.push_back(++c);
  for(int i = 0; i < (int)l.size(); ++i) {
    int d = ++c;
    for(int j = 0; j < l[i]; ++j) { r.push_back(++c); }
    r.push_back(d);
  }
  r.push_back(++c);
}

int main() {
  cin >> N;
  for(int i = 0; i < N - 1; ++i) {
    int v, w; cin >> v >> w; --v; --w; A[v].push_back(w); A[w].push_back(v);
  }
  if(N == 2) {
    cout << "1 2" << endl;
  } else {
    bool ok = false;
    for(int v = 0; v < N; ++v) {
      if((int)A[v].size() != 1) { ok = see(v, -1); break; }
    }
    if(!ok) cout << -1 << endl;
    else {
      go(S, -1);
      vector<int> R1, R2;
      make(L, R1); reverse(L.begin(), L.end()); make(L, R2);
      vector<int> R = min(R1, R2);
      for(int i = 0; i < N; ++i) {
        cout << (i == 0 ? "" : " ") << R[i];
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int N;
vector<int> A[100000];
int S;
vector<int> L;

bool see(int v, int par) {
  int count = 0;
  for(int w : A[v]) {
    if(w != par && (int)A[w].size() != 1) {
      ++count;
      if(!see(w, v)) return false;
    }
  }
  if(par >= 0 && count > 1) return false;
  else if(par < 0 && count > 2) return false;
  else {
    if(count == 0) S = v;
    return true;
  }
}

void go(int v, int par) {
  L.push_back(A[v].size() - 2);
  for(int w : A[v]) {
    if(w != par && (int)A[w].size() != 1) { go(w, v); return; }
  }
}

void make(const vector<int>& l, vector<int>& r) {
  int c = 0;
  r.push_back(++c);
  for(int i = 0; i < (int)l.size(); ++i) {
    int d = ++c;
    for(int j = 0; j < l[i]; ++j) { r.push_back(++c); }
    r.push_back(d);
  }
  r.push_back(++c);
}

int main() {
  cin >> N;
  for(int i = 0; i < N - 1; ++i) {
    int v, w; cin >> v >> w; --v; --w; A[v].push_back(w); A[w].push_back(v);
  }
  if(N == 2) {
    cout << "1 2" << endl;
  } else {
    bool ok = false;
    for(int v = 0; v < N; ++v) {
      if((int)A[v].size() != 1) { ok = see(v, -1); break; }
    }
    if(!ok) cout << -1 << endl;
    else {
      go(S, -1);
      vector<int> R1, R2;
      make(L, R1); reverse(L.begin(), L.end()); make(L, R2);
      vector<int> R = min(R1, R2);
      for(int i = 0; i < N; ++i) {
        cout << (i == 0 ? "" : " ") << R[i];
      }
      cout << endl;
    }
  }
  return 0;
}

```