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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

vector<int> g[100005];
int d[100005];

void dfs(int x, int p) {
    d[x] = d[p] + 1;
    for (int i : g[x]) if (i != p) dfs(i, x);
}

int main() {
    ios::sync_with_stdio(false);
    int n, u, v, c = 0;
    vector<int> a, b, e, z, y;
    cin >> n;
    for (int i = 1; i < n; i++) cin >> u >> v, g[u].push_back(v), g[v].push_back(u);
    u = v = 0;
    dfs(1, 1);
    for (int i = 1; i <= n; i++) if (d[i] > d[u]) u = i;
    for (int i = 1; i <= n; i++) d[i] = 0;
    dfs(u, u);
    for (int i = 1; i <= n; i++) if (d[i] > d[v]) v = i;
    a.push_back(v);
    while (v != u) {
        for (int i : g[v]) if (d[i] + 1 == d[v]) {
            v = i;
            a.push_back(v);
            break;
        }
    }
    for (int i : a) {
        if (i == a[0] || i == a.back()) b.push_back((int)g[i].size());
        else b.push_back((int)g[i].size() - 1);
        c += b.back();
    }
    e = b;
    reverse(e.begin(), e.end());
    if (c != n) {
        cout << -1;
    } else {
        z.resize(n);
        y.resize(n);
        c = 0;
        for (int i : b) {
            z[c + i - 1] = c + 1;
            for (int j = c + 1; j < c + i; j++) z[j - 1] = j + 1;
            c += i;
        }
        c = 0;
        for (int i : e) {
            y[c + i - 1] = c + 1;
            for (int j = c + 1; j < c + i; j++) y[j - 1] = j + 1;
            c += i;
        }
        if (z > y) swap(z, y);
        for (int i : z) cout << i << ' ';
    }
}
