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;

pair<int, int> dfs(int i, int p, const vector<vector<int> > &g) {
	pair<int, int> r(0, i);
	for (int j : g[i]) if (j != p) {
		auto p = dfs(j, i, g);
		r = max(r, make_pair(p.first + 1, p.second));
	}
	return r;
}

bool getPath(int i, int t, int p, const vector<vector<int> > &g, vector<int> &path) {
	if (i == t) {
		path.push_back(i);
		return true;
	}
	for (int j : g[i]) if (j != p) {
		if (getPath(j, t, i, g, path)) {
			path.push_back(i);
			return true;
		}
	}
	return false;
}

int main() {
	int n;
	while (~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> path;
		{
			int root = dfs(0, -1, g).second;
			int target = dfs(root, -1, g).second;
			getPath(root, target, -1, g, path);
		}
		vector<int> ans;
		for (int d = 0; d < 2; ++d) {
			reverse(path.begin(), path.end());
			vector<pair<int, int>> rank;
			vector<bool> vis(n);
			for (int u : path) vis[u] = true;
			int k = 0;
			for (int u : path) {
				rank.emplace_back(k+1, (int)rank.size());
				for (int v : g[u]) if (!vis[v]) {
					rank.emplace_back(k, (int)rank.size());
				}
				++k;
			}
			if (rank.size() != n) {
				continue;
			}
			sort(rank.begin(), rank.end());
			vector<int> v(n);
			for (int i = 0; i < n; ++i)
				v[i] = rank[i].second + 1;
			if (ans.empty() || ans > v)
				ans = v;
		}
		if (ans.empty()) {
			puts("-1");
			continue;
		}
		for (int i = 0; i < (int)ans.size(); ++i) {
			if (i != 0) putchar(' ');
			printf("%d", ans[i]);
		}
		puts("");
	}
}
