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;

const int N = 100005;

int n; 
int ok = 1;
vector <int> g[N];
vector <int> child;

void dfs(int at, int par){
	int cnt = 0, ch = 0;
	for(auto u : g[at]){
		if(u == par) continue;
		if(g[u].size() > 1){
			dfs(u, at);
			++cnt;
		}
		++ch;
	}
	child.push_back(ch);
	ok &= (cnt <= 1);
}

pair <int, int> mx = {-1, 0};

void dfs0(int at, int par, int dep){
	mx = max(mx, {dep, at});
	for(auto u : g[at]){
		if(u == par) continue;
		dfs0(u, at, dep + 1);
	}
}

vector <int> ans;

void check(int S){
	ok = 1;
	child.clear();
	dfs(S, 0);
	reverse(child.begin(), child.end());
	if(!ok) return;
	vector <int> alt;
	int cur = 1;
	for(auto u : child){
		for(int i = cur + 1; i < cur + u; ++i){
			alt.push_back(i);
		}
		alt.push_back(cur);
		cur += u;
	}
	alt.push_back(n);
	if(ans.empty()) ans = alt;
	else ans = min(ans, alt);
}

int main(){
	int x, y;
	scanf("%d", &n);
	for(int i = 2; i <= n; ++i){
		scanf("%d %d", &x, &y);
		g[x].push_back(y);
		g[y].push_back(x);
	}

	dfs0(1, 0, 0);
	int a = mx.second;
	mx = {-1, 0};
	dfs0(a, 0, 0);
	int b = mx.second;

	check(a);
	check(b);
	if(!ans.empty()){
		for(auto u : ans) printf("%d ", u);
		printf("\n");
	}
	else printf("-1\n");
	return 0;
}
