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;
int n;
const int N = 300000;
int tot;
vector <int> bi[N];
int dep[N], A[N], B[N], fa[N], vis[N];
int dfs1(int t, int f = -1, int d = 0)
{
    int v = t;
    dep[t] = d; fa[t] = f;
    for (int i = 0; i < bi[t].size(); ++ i)
        if (bi[t][i] != f)
        {
            int w = dfs1(bi[t][i], t, d + 1);
            if (dep[w] > dep[v]) v = w;
        }
    return v;
}
void dfs2(int t, int d, int A[], int f = -1)
{
    int c = bi[t].size() + (f == -1 || t == d) - 2;
    A[tot + c] = tot;
    for (int i = 1; i <= c; ++ i) A[tot + i - 1] = tot + i;
    tot += c + 1;
    for (int i = 0; i < bi[t].size(); ++ i)
        if (vis[bi[t][i]] && bi[t][i] != f)
            dfs2(bi[t][i], d, A, t);
}

void solve(int pa, int pb, int A[])
{
    tot = 1;
    dfs2(pa, pb, A);
}
int *ans;
int main()
{
    ios::sync_with_stdio();
    cin >> n;
    for (int i = 1; i < n; ++ i)
    {
        int a, b;
        cin >> a >> b;
        bi[a].push_back(b);
        bi[b].push_back(a);
    }

    int pa = dfs1(1), pb = dfs1(pa);
    for (int p = pb; p != -1; p = fa[p]) vis[p] = 1;
    for (int i = 1; i <= n; ++ i) if (!vis[i] && bi[i].size() != 1)
    {
        cout << -1 << endl;
        return 0;
    }
    solve(pa, pb, A); solve(pb, pa, B);
    ans = A;
    for (int i = 1; i <= n; ++ i)
        if (A[i] != B[i])
        {
            if (A[i] > B[i]) ans = B;
            else ans = A;
            break;
        }
    for (int i = 1; i <= n; ++ i) cout << ans[i] << " "; cout << endl;
}
