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

int n;
const int nmax = 100000;
int v[nmax-1];
int w[nmax-1];

vector<int> output(const vector<int> &cost)
{
    vector<int> res;
    int now = 1;
    for (auto v : cost)
    {
        for (int i=1; i<v; i++)
        {
            res.push_back(now + i);
        }
        res.push_back(now);
        now += v;
    }
    return res;
}

int main()
{
    cin >> n;
    if (n == 2)
    {
        cout << "1 2" << endl;
        return 0;
    }

    vector<vector<int> > To(n);
    for (int i = 0; i < n - 1; i++)
    {
        cin >> v[i] >> w[i];
        v[i]--;
        w[i]--;
        To[v[i]].emplace_back(w[i]);
        To[w[i]].emplace_back(v[i]);
    }

    vector<int> score(n), isEnd(n);
    for (int i = 0; i < n; i++)
    {
        if (To[i].size() == 1)
        {
            score[To[i][0]]++;
            isEnd[i] = 1;
        }
    }

    int startp = -1;
    for (int i = 0; i < n; i++)
    {
        if (To[i].size() - score[i] > 2)
        {
            cout << "-1\n";
            return 0;
        }
        if (To[i].size() - score[i] <= 1 && !isEnd[i]) startp = i;
    }

    vector<int> costVec;
    while (1)
    {
        costVec.push_back(score[startp] + 1);
        int nextp = -1;
        isEnd[startp] = 1;
        for (auto v : To[startp]) if (!isEnd[v]) nextp = v;
        if (nextp < 0) break;
        startp = nextp;
    }

    costVec[0]--;
    costVec.insert(costVec.begin(), 1);
    costVec.back()--;
    costVec.push_back(1);

    vector<int> res1 = output(costVec);
    reverse(costVec.begin(), costVec.end());
    vector<int> res2 = output(costVec);
    int flg = -1;
    for (int i=0; i<res1.size(); i++)
    {
        if (flg == -1)
        {
            if (res2[i] < res1[i]) flg = 2;
            if (res2[i] > res1[i]) flg = 1;
        }
    }
    for (int i=0; i<res1.size(); i++) cout << (flg==2 ? res2[i] : res1[i]) << " ";
    cout << endl;
}

### Prompt

Please formulate 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;

int n;
const int nmax = 100000;
int v[nmax-1];
int w[nmax-1];

vector<int> output(const vector<int> &cost)
{
    vector<int> res;
    int now = 1;
    for (auto v : cost)
    {
        for (int i=1; i<v; i++)
        {
            res.push_back(now + i);
        }
        res.push_back(now);
        now += v;
    }
    return res;
}

int main()
{
    cin >> n;
    if (n == 2)
    {
        cout << "1 2" << endl;
        return 0;
    }

    vector<vector<int> > To(n);
    for (int i = 0; i < n - 1; i++)
    {
        cin >> v[i] >> w[i];
        v[i]--;
        w[i]--;
        To[v[i]].emplace_back(w[i]);
        To[w[i]].emplace_back(v[i]);
    }

    vector<int> score(n), isEnd(n);
    for (int i = 0; i < n; i++)
    {
        if (To[i].size() == 1)
        {
            score[To[i][0]]++;
            isEnd[i] = 1;
        }
    }

    int startp = -1;
    for (int i = 0; i < n; i++)
    {
        if (To[i].size() - score[i] > 2)
        {
            cout << "-1\n";
            return 0;
        }
        if (To[i].size() - score[i] <= 1 && !isEnd[i]) startp = i;
    }

    vector<int> costVec;
    while (1)
    {
        costVec.push_back(score[startp] + 1);
        int nextp = -1;
        isEnd[startp] = 1;
        for (auto v : To[startp]) if (!isEnd[v]) nextp = v;
        if (nextp < 0) break;
        startp = nextp;
    }

    costVec[0]--;
    costVec.insert(costVec.begin(), 1);
    costVec.back()--;
    costVec.push_back(1);

    vector<int> res1 = output(costVec);
    reverse(costVec.begin(), costVec.end());
    vector<int> res2 = output(costVec);
    int flg = -1;
    for (int i=0; i<res1.size(); i++)
    {
        if (flg == -1)
        {
            if (res2[i] < res1[i]) flg = 2;
            if (res2[i] > res1[i]) flg = 1;
        }
    }
    for (int i=0; i<res1.size(); i++) cout << (flg==2 ? res2[i] : res1[i]) << " ";
    cout << endl;
}

```