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

const int N = 100010;
int n , ans[2][N] , vis[N];
vector<int> to[N];

inline int read() {
    int x = 0 ; char c = getchar(); bool f = 0;
    while(c < '0' || c > '9') f |= (c=='-') , c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + (c^48) , c = getchar();
    return f ? -x : x;
}

int dep[N] , f[N];
void DFS(int x , int fa) {
    dep[x] = dep[f[x] = fa] + 1;
    for(auto &y : to[x]) if(y != fa) DFS(y , x);
}

int main() {
    n = read();
    for(int i = 1 ; i < n ; ++i) {
        int x = read() , y = read();
        to[x].push_back(y) , to[y].push_back(x);
    }
    int rt1 , rt2;
    DFS(1 , 0);
    rt1 = max_element(dep + 1 , dep + 1 + n) - dep;
    DFS(rt1 , 0);
    rt2 = max_element(dep + 1 , dep + 1 + n) - dep;
    memset(vis , -1 , sizeof(vis));
    for(int x = rt2 ; x ; x = f[x]) vis[x] = 0;
    for(int i = 1 ; i <= n ; ++i) if(vis[i] == -1) {
        int t = -1;
        for(auto y : to[i]) if(vis[y] != -1) {
            t = y ; break;
        }
        if(t == -1) return puts("-1") & 0;
        vis[t]++;
    }
    for(int k = 0 ; k < 2 ; ++k) {
        int x = k==0 ? rt1 : rt2 , cnt = 1 , nxt = 0, las = 0;
        while(x) {
            for(int i = 1 ; i <= vis[x] ; ++i) ans[k][cnt + i - 1] = cnt + i;
            ans[k][cnt + vis[x]] = cnt , cnt += vis[x] + 1;
            
            nxt = 0;
            for(auto y : to[x]) if(vis[y] != -1 && y != las) {nxt = y ; break;}
            las = x , x = nxt;
        }
    }
    int k = (memcmp(ans[0]+1 , ans[1]+1 , n*sizeof(int)) < 0) ? 0 : 1;
    for(int i = 1 ; i <= n ; ++i) printf("%d " , ans[k][i]);
    return 0;
}
### Prompt

Your task is to create 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 <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 100010;
int n , ans[2][N] , vis[N];
vector<int> to[N];

inline int read() {
    int x = 0 ; char c = getchar(); bool f = 0;
    while(c < '0' || c > '9') f |= (c=='-') , c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + (c^48) , c = getchar();
    return f ? -x : x;
}

int dep[N] , f[N];
void DFS(int x , int fa) {
    dep[x] = dep[f[x] = fa] + 1;
    for(auto &y : to[x]) if(y != fa) DFS(y , x);
}

int main() {
    n = read();
    for(int i = 1 ; i < n ; ++i) {
        int x = read() , y = read();
        to[x].push_back(y) , to[y].push_back(x);
    }
    int rt1 , rt2;
    DFS(1 , 0);
    rt1 = max_element(dep + 1 , dep + 1 + n) - dep;
    DFS(rt1 , 0);
    rt2 = max_element(dep + 1 , dep + 1 + n) - dep;
    memset(vis , -1 , sizeof(vis));
    for(int x = rt2 ; x ; x = f[x]) vis[x] = 0;
    for(int i = 1 ; i <= n ; ++i) if(vis[i] == -1) {
        int t = -1;
        for(auto y : to[i]) if(vis[y] != -1) {
            t = y ; break;
        }
        if(t == -1) return puts("-1") & 0;
        vis[t]++;
    }
    for(int k = 0 ; k < 2 ; ++k) {
        int x = k==0 ? rt1 : rt2 , cnt = 1 , nxt = 0, las = 0;
        while(x) {
            for(int i = 1 ; i <= vis[x] ; ++i) ans[k][cnt + i - 1] = cnt + i;
            ans[k][cnt + vis[x]] = cnt , cnt += vis[x] + 1;
            
            nxt = 0;
            for(auto y : to[x]) if(vis[y] != -1 && y != las) {nxt = y ; break;}
            las = x , x = nxt;
        }
    }
    int k = (memcmp(ans[0]+1 , ans[1]+1 , n*sizeof(int)) < 0) ? 0 : 1;
    for(int i = 1 ; i <= n ; ++i) printf("%d " , ans[k][i]);
    return 0;
}
```