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
#include <bits/stdc++.h>

using namespace std;

const int Nmax = 1e5 + 5;

vector<int> path, v[Nmax], p1, p2;
int x, y, i, n, dist[Nmax];

void dfs(int node)
{
    for(auto it : v[node])
        if(!dist[it])
        {
            dist[it] = dist[node] + 1;
            dfs(it);
        }
}

int farthest(int node)
{
    memset(dist, 0, sizeof(dist));
    dist[node] = 1;
    dfs(node);
    int i, ans = 0;
    for(i=1; i<=n; ++i)
        if(dist[i] > dist[ans]) ans = i;
    return ans;
}

bool find_path(int node, int x, int dad = 0)
{
    bool ok = (node == x);
    for(auto it : v[node])
        if(it != dad)
            if(find_path(it, x, node))
                ok = 1;
    if(ok) path.push_back(node);
    return ok;
}

vector<int> solve()
{
    vector<int> ans;
    int i, j, p = 1;

    for(i=0; i<path.size(); ++i)
    {
        for(j=1; j<=path[i]; ++j)
            ans.push_back(p + j);
        ans.push_back(p);
        p += path[i] + 1;
    }
    return ans;
}

int main()
{
  //  freopen("input", "r", stdin);
  //  freopen("output", "w", stdout);

    cin >> n;
    for(i=1; i<n; ++i)
    {
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
    }

    x = farthest(1);
    y = farthest(x);
    find_path(x, y);

    for(i=1; i<path.size()-1; ++i)
        path[i] = v[path[i]].size() - 2;
    path.front() = path.back() = 0;

    int sum = path.size();
    for(auto node : path) sum += node;

    if(sum != n)
    {
        cout << "-1\n";
        return 0;
    }

    p1 = solve();
    reverse(path.begin(), path.end());
    p2 = solve();

    for(i=0; i<n; ++i)
        if(p1[i] > p2[i])
        {
            swap(p1, p2);
            break;
        }
        else if(p1[i] < p2[i]) break;

    for(auto val : p1) cout << val << ' ';
    return 0;
}
