#include <bits/stdc++.h>
using namespace std;

using ll = long long;

const int N = 100005;

int n, d[N], c[N], p[N], q, t;
vector<int> e[N], r, s;

void f(int x, int y){
    int u = int(e[x].size()) - d[x];
    if(d[x] == 1){
        if(!q){
            p[1] = q++;
            for(int i = t + 2; i <= t + u; i++) p[i] = q++;
            p[2] = q++;
            t += (u + 1); 
        }
        else{
            for(int i = t + 1; i < t + u; i++) p[i] = q++;
            p[t] = q++;
            p[n] = q++;
        }
    }
    else{
        for(int i = t + 1; i <= t + u; i++) p[i] = q++;
        p[t] = q++;
        t += (u + 1);
    }
    for(int i : e[x]) if(i != y && d[i]) f(i, x);
}

int main(){
    scanf("%d", &n);
    if(n == 1){ puts("1"); return 0; }
    for(int i = 0, x, y; i < n - 1; i++){
        scanf("%d%d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    
    for(int i = 1; i <= n; i++) c[i] = (e[i].size() > 1);
    for(int i = 1; i <= n; i++) for(int j : e[i]) if(c[i] && c[j]) d[i]++;
    if(count(d + 1, d + n + 1, 1) > 2){ puts("-1"); return 0; }
    
    if(count(d + 1, d + n + 1, 1) == 0){
        printf("1 ");
        for(int i = 3; i <= n - 1; i++) printf("%d ", i);
        printf("2 %d", n);
        return 0;
    }

    q = 0; t = 1;
    for(int i = 1; i <= n; i++) if(d[i] == 1){ f(i, 0); break; }
    r.resize(n);
    for(int i = 1; i <= n; i++) r[p[i]] = i;
    
    q = 0; t = 1;
    for(int i = n; i >= 1; i--) if(d[i] == 1){ f(i, 0); break; }
    s.resize(n);
    for(int i = 1; i <= n; i++) s[p[i]] = i;
    
    r = min(r, s);
    for(int i : r) printf("%d ", i);
    puts("");
}

### 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 namespace std;

using ll = long long;

const int N = 100005;

int n, d[N], c[N], p[N], q, t;
vector<int> e[N], r, s;

void f(int x, int y){
    int u = int(e[x].size()) - d[x];
    if(d[x] == 1){
        if(!q){
            p[1] = q++;
            for(int i = t + 2; i <= t + u; i++) p[i] = q++;
            p[2] = q++;
            t += (u + 1); 
        }
        else{
            for(int i = t + 1; i < t + u; i++) p[i] = q++;
            p[t] = q++;
            p[n] = q++;
        }
    }
    else{
        for(int i = t + 1; i <= t + u; i++) p[i] = q++;
        p[t] = q++;
        t += (u + 1);
    }
    for(int i : e[x]) if(i != y && d[i]) f(i, x);
}

int main(){
    scanf("%d", &n);
    if(n == 1){ puts("1"); return 0; }
    for(int i = 0, x, y; i < n - 1; i++){
        scanf("%d%d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    
    for(int i = 1; i <= n; i++) c[i] = (e[i].size() > 1);
    for(int i = 1; i <= n; i++) for(int j : e[i]) if(c[i] && c[j]) d[i]++;
    if(count(d + 1, d + n + 1, 1) > 2){ puts("-1"); return 0; }
    
    if(count(d + 1, d + n + 1, 1) == 0){
        printf("1 ");
        for(int i = 3; i <= n - 1; i++) printf("%d ", i);
        printf("2 %d", n);
        return 0;
    }

    q = 0; t = 1;
    for(int i = 1; i <= n; i++) if(d[i] == 1){ f(i, 0); break; }
    r.resize(n);
    for(int i = 1; i <= n; i++) r[p[i]] = i;
    
    q = 0; t = 1;
    for(int i = n; i >= 1; i--) if(d[i] == 1){ f(i, 0); break; }
    s.resize(n);
    for(int i = 1; i <= n; i++) s[p[i]] = i;
    
    r = min(r, s);
    for(int i : r) printf("%d ", i);
    puts("");
}

```