#include <bits/stdc++.h>

using std::cerr;
using std::endl;

inline int rd() {
  int x; scanf("%d", &x);
  return x;
}

const int N = 1e5 + 233;

int n, dis[N], tag[N], deg[N], stk[N], top, A[N], B[N];
std::vector<int> G[N];

int dfs(int x, int fa) {
  dis[x] = dis[fa] + 1;
  int ret = 0;
  for (int y : G[x]) if (y != fa) {
    int z = dfs(y, x);
    if (dis[z] > dis[ret])
      ret = z;
  }
  if (!ret) return x;
  return ret;
}

int find_path(int x, int fa, int t) {
  if (x == t) {
    tag[x] = 1;
    stk[++top] = x;
    return 1;
  }
  for (int y : G[x]) if (y != fa) {
    if (find_path(y, x, t)) {
      tag[x] = 1;
      stk[++top] = x;
      return 1;
    }
  }
  return 0;
}

inline void work(int st, int f[]) {
  if (stk[1] != st)
    std::reverse(stk + 1, stk + top + 1);
  int sum = 0, w = 0;
  for (int i = 1; i <= top; ++i) {
    int x = stk[i], s = G[x].size() - 1;
    if (i != 1 && i != top) s -= 1;
    for (int j = 1; j <= s; ++j)
      f[++w] = sum + j + 1;
    f[++w] = sum + 1;
    sum += s + 1;
  }
}

int main() {
  /// freopen("data", "r", stdin);
  n = rd();
  for (int i = 1; i < n; ++i) {
    int x = rd(), y = rd();
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int x = dfs(1, 0), y = dfs(x, 0);
  find_path(x, 0, y);
  for (int x = 1; x <= n; ++x) {
    if (tag[x]) continue;
    int ok = 0;
    for (int y : G[x])
      if (tag[y]) {
        ok = 1; break;
      }
    if (!ok)
      return puts("-1"), 0;
  }
  work(x, A), work(y, B);
  for (int i = 1; i <= n; ++i) {
    if (A[i] < B[i] || i == n) {
      for (int j = 1; j <= n; ++j)
        printf("%d ", A[j]);
      break;
    }
    if (A[i] > B[i]) {
      for (int j = 1; j <= n; ++j)
        printf("%d ", B[j]);
      break;
    }
  }
  putchar('\n');
  return 0;
}


### Prompt

Your challenge is to write a CPP solution 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 std::cerr;
using std::endl;

inline int rd() {
  int x; scanf("%d", &x);
  return x;
}

const int N = 1e5 + 233;

int n, dis[N], tag[N], deg[N], stk[N], top, A[N], B[N];
std::vector<int> G[N];

int dfs(int x, int fa) {
  dis[x] = dis[fa] + 1;
  int ret = 0;
  for (int y : G[x]) if (y != fa) {
    int z = dfs(y, x);
    if (dis[z] > dis[ret])
      ret = z;
  }
  if (!ret) return x;
  return ret;
}

int find_path(int x, int fa, int t) {
  if (x == t) {
    tag[x] = 1;
    stk[++top] = x;
    return 1;
  }
  for (int y : G[x]) if (y != fa) {
    if (find_path(y, x, t)) {
      tag[x] = 1;
      stk[++top] = x;
      return 1;
    }
  }
  return 0;
}

inline void work(int st, int f[]) {
  if (stk[1] != st)
    std::reverse(stk + 1, stk + top + 1);
  int sum = 0, w = 0;
  for (int i = 1; i <= top; ++i) {
    int x = stk[i], s = G[x].size() - 1;
    if (i != 1 && i != top) s -= 1;
    for (int j = 1; j <= s; ++j)
      f[++w] = sum + j + 1;
    f[++w] = sum + 1;
    sum += s + 1;
  }
}

int main() {
  /// freopen("data", "r", stdin);
  n = rd();
  for (int i = 1; i < n; ++i) {
    int x = rd(), y = rd();
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int x = dfs(1, 0), y = dfs(x, 0);
  find_path(x, 0, y);
  for (int x = 1; x <= n; ++x) {
    if (tag[x]) continue;
    int ok = 0;
    for (int y : G[x])
      if (tag[y]) {
        ok = 1; break;
      }
    if (!ok)
      return puts("-1"), 0;
  }
  work(x, A), work(y, B);
  for (int i = 1; i <= n; ++i) {
    if (A[i] < B[i] || i == n) {
      for (int j = 1; j <= n; ++j)
        printf("%d ", A[j]);
      break;
    }
    if (A[i] > B[i]) {
      for (int j = 1; j <= n; ++j)
        printf("%d ", B[j]);
      break;
    }
  }
  putchar('\n');
  return 0;
}


```