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;
class PermutationTree {
private:
    int n;
    vector<vector<int>> edge;
    vector<int> parent, dep;
    vector<bool> isOnDiameter;
    int bfs(int s) {
        parent.assign(n, -1);
        dep.assign(n, 0);
        queue<int> que;
        que.push(s);
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            for (int v : edge[u]) {
                if (v == parent[u])
                    continue;
                parent[v] = u;
                dep[v] = dep[u] + 1;
                que.push(v);
            }
        }
        return max_element(dep.begin(), dep.end()) - dep.begin();
    }
    vector<int> findPermutation(const vector<int> &k) {
        vector<int> perm;
        int cnt = 0;
        for (int x : k) {
            int u = cnt++;
            for (int i = 0; i < x; ++i)
                perm.push_back(cnt++);
            perm.push_back(u);
        }
        return perm;
    }
public:
    void solve() {
        cin >> n;
        edge.resize(n);
        isOnDiameter.resize(n);
        for (int i = 0; i < n - 1; ++i) {
            int u, v;
            cin >> u >> v;
            --u;
            --v;
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
        int s = bfs(0);
        int t = bfs(s);
        vector<int> k;
        for (int u = t; u != -1; u = parent[u]) {
            isOnDiameter[u] = true;
            k.push_back(u == s || u == t ? 0 : (int)edge[u].size() - 2);
        }
        for (int i = 0; i < n; ++i) {
            if (!isOnDiameter[i] && (int)edge[i].size() != 1) {
                cout << -1 << endl;
                return;
            }
        }
        auto ans = findPermutation(k);
        reverse(k.begin(), k.end());
        ans = min(ans, findPermutation(k));
        for (int i : ans)
            cout << i + 1 << " \n"[i == ans.back()];
    }
};
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    PermutationTree().solve();
    return 0;
}