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 main() {
    int n; scanf("%d", &n);
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; ++i) {
        int u, v; scanf("%d%d", &u, &v);
        u--, v--;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    vector<int> dep(n), fa(n);
    function<void(int, int)> Dfs = [&](int x, int p) {
        dep[x] = ~p ? dep[p] + 1 : 0;
        fa[x] = p;
        for (int u : g[x]) {
            if (u == p) continue;
            Dfs(u, x);
        }
    };

    Dfs(0, -1);
    int x = max_element(dep.begin(), dep.end()) - dep.begin();
    Dfs(x, -1);
    int y = max_element(dep.begin(), dep.end()) - dep.begin();
    int z = y, cnt = 0;
    while (z != -1) {
        cnt += g[z].size();
        z = fa[z];
    }
    cnt -= dep[y] - 1;
    if (cnt != n) {
        puts("-1");
        return 0;
    }

    vector<int> v;
    function<void(int, int, int)> Solve = [&](int x, int p, int k) {
        int ch = 0;
        for (int u : g[x]) ch += (u != p);
        for (int i = 1; i <= ch - 1; ++i) v.push_back(k + i);
        v.push_back(k);
        int cc = 0;
        for (int u : g[x]) {
            if (u == p) continue;
            if (g[u].size() > 1) {
                assert(cc++ == 0);
                Solve(u, x, k + ch);
            }
        }
        if (cc == 0) v.push_back(k + ch);
    };

    Solve(x, -1, 1);
    auto ans = v;
    v.clear();
    Solve(y, -1, 1);
    ans = min(ans, v);
    for (int u : ans) printf("%d ", u);
    puts("");
}
