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;

const int N = 100233;
int n, u, v, p, dep[N], vis[N], size[N];
vector<int> G[N], t(1, 1);
void dfs(int x, int pre) {
  dep[x] = dep[pre] + 1;
  for (auto y : G[x]) if (y != pre) {
    dfs(y, x);
  }
}
bool dfs(int x, int pre, int g) {
  vis[x] = ++p;
  if (x == g) return true;
  for (auto y : G[x]) if (y != pre) {
    if (dfs(y, x, g)) return true;
  }
  vis[x] = 0;
  --p;
  return false;
}
void get() {
  vector<int> f(1, 0);
  for (int i = 1, a = 0; i <= p; i++) {
    for (int j = 1; j <= size[i]; j++) f.push_back(a + j + 1);
    f.push_back(a + 1);
    a += size[i] + 1;
  }
  t = min(t, f);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) if (dep[i] > dep[u]) {
    u = i;
  }
  dfs(u, 0);
  for (int i = 1; i <= n; i++) if (dep[i] > dep[v]) {
    v = i;
  }
  dfs(u, 0, v);
  for (int i = 1; i <= n; i++) if (!vis[i]) {
    if (G[i].size() > 1) return puts("-1"), 0;
    size[vis[G[i][0]]]++;
  }
  get();
  reverse(size + 1, size + 1 + p);
  get();
  for (int i = 1; i <= n; i++) {
    printf("%d%c", t[i], i == n ? '\n' : ' ');
  }
}
