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 <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 1e5;

int n;
vector<int> g[maxn];
int size[maxn], deg[maxn];

vector<int> path;
bool fail;

void Dfs1(int u, int p) {
  path.push_back(size[u]);
  int cnt = 0;
  for (int v : g[u]) {
    if (v != p && size[v] >= 0) {
      ++cnt;
      Dfs1(v, u);
    }
  }
  if (cnt >= 2) fail = true;
}

void Exit(void) {
  puts("-1");
  exit(0);
}

int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u; --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  if (n == 2) {
    puts("1 2");
    return 0;
  }

  for (int i = 0; i < n; ++i) {
    if (g[i].size() == 1) {
      size[i] = -1;
      int j = g[i][0];
      ++size[j];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j : g[i]) {
      if (size[i] >= 0 && size[j] >= 0) {
        ++deg[i];
      }
    }
  }

  int start = -1;
  for (int i = 0; i < n; ++i) {
    if (size[i] >= 0 && deg[i] <= 1) {
      start = i;
      break;
    }
  }
  if (start == -1) throw;
  fail = false;
  path.clear();
  Dfs1(start, -1);
  if (fail) Exit();

  int cnt_cent = 0;
  for (int i = 0; i < n; ++i) {
    cnt_cent += (size[i] >= 0);
  }
  if (path.size() != cnt_cent) Exit();

  if (path[0] >= 1) {
    --path[0];
    path.insert(path.begin(), 0);
  }
  if (path.back() >= 1) {
    --path.back();
    path.push_back(0);
  }
  for (int i = 0, j = (int)path.size() - 1; i < path.size(); ++i, --j) {
    if (path[i] != path[j]) {
      if (path[i] > path[j]) {
        reverse(path.begin(), path.end());
      }
      break;
    }
  }

  int val = 1;
  for (int a : path) {
    for (int i = 0; i < a; ++i) {
      printf("%d ", val + i + 1);
    }
    printf("%d ", val);
    val += a + 1;
  }
  printf("\n");
  return 0;
}
