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>

const int N = 100005;

int n, head[N], tot, deg[N];
struct edge { int to, nxt; } e[N << 1];

void link(int x, int y) {
	e[++tot] = (edge) { y, head[x] }, head[x] = tot, ++deg[x];
	e[++tot] = (edge) { x, head[y] }, head[y] = tot, ++deg[y];
}

int f[N], max, root, p;
bool vis[N];

void dfs(int x, int dep = 0) {
	if (dep >= max) max = dep, p = x;
	for (int i = head[x]; i; i = e[i].nxt)
		if (e[i].to != f[x]) f[e[i].to] = x, dfs(e[i].to, dep + 1);
}

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1, x, y; i < n; ++i)
		std::cin >> x >> y, link(x, y);
	dfs(1), max = 0, root = p; f[p] = 0, dfs(p);
	for (int i = p; i != root; i = f[i]) vis[i] = 1; vis[root] = 1;
	for (int i = f[p]; i != root; i = f[i]) deg[i] -= 2; deg[p] -= 1, deg[root] -= 1;
	for (int i = 1; i <= n; ++i) if (!vis[i] && !vis[f[i]])
		return std::cout << "-1\n", 0;
	std::vector<int> d, ans;
	for (int i = p; i != root; i = f[i]) d.emplace_back(deg[i]); d.emplace_back(0);
	d = std::min(d, std::vector<int>(d.rbegin(), d.rend()));
	int now = 1;
	for (int i = 0; i < d.size(); ++i) {
		for (int j = 0; j < d[i]; ++j) ans.emplace_back(now + j + 1);
		ans.emplace_back(now);
		now += d[i] + 1;
	}
	for (int i = 0; i < n; ++i)
		std::cout << ans[i] << ' ';
	std::cout << '\n';
	return 0;
}