#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
struct data {
  int nt, to;
} a[N << 1];
int head[N], dep[N], f[N], d[N], n, cnt = 0;

void add(int x, int y) {
  a[++cnt].to = y;
  a[cnt].nt = head[x];
  head[x] = cnt;
}

void dfs(int u, int fa) {
  f[u] = fa, dep[u] = dep[fa] + 1;
  for(int i = head[u]; i; i = a[i].nt) {
    int to = a[i].to;
    if(to == fa) continue;
    dfs(to, u);
  }
}

vector<int> g;
int ans = 0, S, T, tag[N];

void GetDia() {
  dfs(1, 0);
  for(int i = 1; i <= n; ++i) if(dep[i] > ans) ans = dep[i], S = i;
  memset(dep, 0, sizeof dep), memset(f, 0, sizeof f);
  ans = 0;
  dfs(S, 0);
  for(int i = 1; i <= n; ++i) if(dep[i] > ans) ans = dep[i], T = i;
  for(int u = T; u; u = f[u]) {
    tag[u] = 1;
//    printf("%d ", u);
    g.push_back(d[u]);
  }
//  puts("");
  for(int i = 1; i <= n; ++i) {
    if(!tag[i]) {
      if(d[i] >= 2) {
        puts("-1");
        exit(0);
      }
    }
  }
}

void go(int a[]) {
  int now = 0;
  for(int i = 0; i < g.size(); ++i) {
    int w = g[i];
    if(i && i != g.size() - 1) w--;
//    printf("%d ", w);
    a[now + 1] = now + w;
    for(int j = 1; j < w; ++j) a[now + 1 + j] = now + j;
    now += w;
  }
//  puts("www");
}

int p[N], p1[N], p2[N];

int main() {
  scanf("%d", &n);
  for(int i = 2, x, y; i <= n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
    d[x]++, d[y]++;
  } 
  GetDia();
  go(p);
  for(int i = 1; i <= n; ++i) p1[p[i]] = i;
  reverse(g.begin(), g.end());
  go(p);
  for(int i = 1; i <= n; ++i) p2[p[i]] = i;
  int now = 1;
  while(now <= n && p1[now] == p2[now]) ++now;
  if(p1[now] < p2[now]) 
    for(int i = 1; i <= n; ++i) printf("%d ", p1[i]);
  else 
    for(int i = 1; i <= n; ++i) printf("%d ", p2[i]);
  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;
const int N = 1e5 + 10;
struct data {
  int nt, to;
} a[N << 1];
int head[N], dep[N], f[N], d[N], n, cnt = 0;

void add(int x, int y) {
  a[++cnt].to = y;
  a[cnt].nt = head[x];
  head[x] = cnt;
}

void dfs(int u, int fa) {
  f[u] = fa, dep[u] = dep[fa] + 1;
  for(int i = head[u]; i; i = a[i].nt) {
    int to = a[i].to;
    if(to == fa) continue;
    dfs(to, u);
  }
}

vector<int> g;
int ans = 0, S, T, tag[N];

void GetDia() {
  dfs(1, 0);
  for(int i = 1; i <= n; ++i) if(dep[i] > ans) ans = dep[i], S = i;
  memset(dep, 0, sizeof dep), memset(f, 0, sizeof f);
  ans = 0;
  dfs(S, 0);
  for(int i = 1; i <= n; ++i) if(dep[i] > ans) ans = dep[i], T = i;
  for(int u = T; u; u = f[u]) {
    tag[u] = 1;
//    printf("%d ", u);
    g.push_back(d[u]);
  }
//  puts("");
  for(int i = 1; i <= n; ++i) {
    if(!tag[i]) {
      if(d[i] >= 2) {
        puts("-1");
        exit(0);
      }
    }
  }
}

void go(int a[]) {
  int now = 0;
  for(int i = 0; i < g.size(); ++i) {
    int w = g[i];
    if(i && i != g.size() - 1) w--;
//    printf("%d ", w);
    a[now + 1] = now + w;
    for(int j = 1; j < w; ++j) a[now + 1 + j] = now + j;
    now += w;
  }
//  puts("www");
}

int p[N], p1[N], p2[N];

int main() {
  scanf("%d", &n);
  for(int i = 2, x, y; i <= n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
    d[x]++, d[y]++;
  } 
  GetDia();
  go(p);
  for(int i = 1; i <= n; ++i) p1[p[i]] = i;
  reverse(g.begin(), g.end());
  go(p);
  for(int i = 1; i <= n; ++i) p2[p[i]] = i;
  int now = 1;
  while(now <= n && p1[now] == p2[now]) ++now;
  if(p1[now] < p2[now]) 
    for(int i = 1; i <= n; ++i) printf("%d ", p1[i]);
  else 
    for(int i = 1; i <= n; ++i) printf("%d ", p2[i]);
  return 0;
}
```