#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>

#define PB push_back

using namespace std;

const int N = 1e5 + 500;

int n,g[5],g2[5],ind,col[N],par[N];
vector < int > v[N], sol[2];

void dfs(int x,int lst,int dis){
    if(g2[ind] < dis){
        g2[ind] = dis;
        g[ind] = x;
    }
    for(int k : v[x])
        if(k != lst)
            dfs(k, x, dis + 1);
}

void root(int x,int lst){
    par[x] = lst;
    for(int k : v[x])
        if(k != lst)
            root(k, x);
}

int main(){
    scanf("%d", &n);
    for(int i = 0;i<n-1;i++){
        int x,y;scanf("%d%d",&x,&y);
        v[x].PB(y);
        v[y].PB(x);
    }
    ind = 0;
    dfs(6,6,0);
    ind = 1;
    dfs(g[0], g[0], 0);
    int st = g[0];
    int en = g[1];
    for(int bla = 0;bla < 2;bla++){
        root(st, 0);
        int cur = en,lst = en;
        int cnt = n;
        while(cur){
            for(int k : v[cur])
                if(k != lst && k != par[cur])
                    col[k] = cnt--;
            col[cur] = cnt--;
            sol[bla].PB(col[cur]);
            for(int k : v[cur])
                if(k != lst && k != par[cur])
                    sol[bla].PB(col[k]);
            lst = cur;
            cur = par[cur];
        }
        reverse(sol[bla].begin(),sol[bla].end());
        if(cnt != 0){
            printf("-1\n");
            return 0;
        }
        swap(st, en);
    }
    if(sol[0] > sol[1]) swap(sol[0], sol[1]);
    for(int k : sol[0])
        printf("%d ", k);
    printf("\n");
}

### 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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>

#define PB push_back

using namespace std;

const int N = 1e5 + 500;

int n,g[5],g2[5],ind,col[N],par[N];
vector < int > v[N], sol[2];

void dfs(int x,int lst,int dis){
    if(g2[ind] < dis){
        g2[ind] = dis;
        g[ind] = x;
    }
    for(int k : v[x])
        if(k != lst)
            dfs(k, x, dis + 1);
}

void root(int x,int lst){
    par[x] = lst;
    for(int k : v[x])
        if(k != lst)
            root(k, x);
}

int main(){
    scanf("%d", &n);
    for(int i = 0;i<n-1;i++){
        int x,y;scanf("%d%d",&x,&y);
        v[x].PB(y);
        v[y].PB(x);
    }
    ind = 0;
    dfs(6,6,0);
    ind = 1;
    dfs(g[0], g[0], 0);
    int st = g[0];
    int en = g[1];
    for(int bla = 0;bla < 2;bla++){
        root(st, 0);
        int cur = en,lst = en;
        int cnt = n;
        while(cur){
            for(int k : v[cur])
                if(k != lst && k != par[cur])
                    col[k] = cnt--;
            col[cur] = cnt--;
            sol[bla].PB(col[cur]);
            for(int k : v[cur])
                if(k != lst && k != par[cur])
                    sol[bla].PB(col[k]);
            lst = cur;
            cur = par[cur];
        }
        reverse(sol[bla].begin(),sol[bla].end());
        if(cnt != 0){
            printf("-1\n");
            return 0;
        }
        swap(st, en);
    }
    if(sol[0] > sol[1]) swap(sol[0], sol[1]);
    for(int k : sol[0])
        printf("%d ", k);
    printf("\n");
}

```