#include <bits/stdc++.h>

using namespace std;

int n;
vector<int> lis[100100];
int deg[100100];
vector<int> sco;

void dfs(int here, int p) {
    sco.push_back(deg[here]-2);
    for (int &there :lis[here]) {
        if (deg[there]==1||there==p) continue;
        dfs(there,here);
    }
}

int main() {
    int i;

    scanf("%d",&n);
    for (i=0;i<n-1;i++) {
        int a, b;
        scanf("%d%d",&a,&b);--a;--b;
        lis[a].push_back(b);
        lis[b].push_back(a);
        deg[a]++; deg[b]++;
    }
    if (n==2) {
        printf("1 2\n");
        return 0;
    }
    int st = -1;
    for (i=0;i<n;i++) {
        if (deg[i]==1) continue;
        int cnt = 0;
        for (int &v: lis[i]) if (deg[v]!=1) cnt++;
        if (!cnt) {
            printf("1 ");
            for (int j=3;j<n;j++) printf("%d ",j);
            printf("2 %d\n",n);
            return 0;
        }
        if (cnt>=3) {
            printf("-1\n");
            return 0;
        }
        if (cnt==1) st = i;
    }
    dfs(st,-1);
    int t = 0;
    for (i=0;i<sco.size();i++) {
        if (sco[i]!=sco[(int)sco.size()-1-i]) {
            if (sco[i]<sco[(int)sco.size()-1-i]) t = 0;
            else t = 1;
            break;
        }
    }
    if (t) reverse(sco.begin(),sco.end());
    int p = 1;
    printf("%d ",p); p++;
    for (i=0;i<sco.size();i++) {
        for (int j=0;j<sco[i];j++) printf("%d ",p+j+1);
        printf("%d ",p); p += sco[i]+1;
    }
    printf("%d\n",n);

    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

int n;
vector<int> lis[100100];
int deg[100100];
vector<int> sco;

void dfs(int here, int p) {
    sco.push_back(deg[here]-2);
    for (int &there :lis[here]) {
        if (deg[there]==1||there==p) continue;
        dfs(there,here);
    }
}

int main() {
    int i;

    scanf("%d",&n);
    for (i=0;i<n-1;i++) {
        int a, b;
        scanf("%d%d",&a,&b);--a;--b;
        lis[a].push_back(b);
        lis[b].push_back(a);
        deg[a]++; deg[b]++;
    }
    if (n==2) {
        printf("1 2\n");
        return 0;
    }
    int st = -1;
    for (i=0;i<n;i++) {
        if (deg[i]==1) continue;
        int cnt = 0;
        for (int &v: lis[i]) if (deg[v]!=1) cnt++;
        if (!cnt) {
            printf("1 ");
            for (int j=3;j<n;j++) printf("%d ",j);
            printf("2 %d\n",n);
            return 0;
        }
        if (cnt>=3) {
            printf("-1\n");
            return 0;
        }
        if (cnt==1) st = i;
    }
    dfs(st,-1);
    int t = 0;
    for (i=0;i<sco.size();i++) {
        if (sco[i]!=sco[(int)sco.size()-1-i]) {
            if (sco[i]<sco[(int)sco.size()-1-i]) t = 0;
            else t = 1;
            break;
        }
    }
    if (t) reverse(sco.begin(),sco.end());
    int p = 1;
    printf("%d ",p); p++;
    for (i=0;i<sco.size();i++) {
        for (int j=0;j<sco[i];j++) printf("%d ",p+j+1);
        printf("%d ",p); p += sco[i]+1;
    }
    printf("%d\n",n);

    return 0;
}

```