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<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

vector<int> G[100100];
int N;

void input(){
	scanf("%d", &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);
	}
}

int dis[100100];
void dfs1(int v, int p){
	for(int i = 0; i < G[v].size(); ++i){
		int u = G[v][i];
		if(u == p) continue;
		dis[u] = dis[v] + 1;
		dfs1(u, v);
	}
}

int sz[100100];
void dfs(int v, int p){
	sz[v] = 1;
	for(int i = 0; i < G[v].size(); ++i){
		int u = G[v][i];
		if(u == p) continue;
		dfs(u, v);
		sz[v] += sz[u];
	}
}

vector<int> perm;
bool dfs3(int v, int p, int cur){
	int cnt = 0;
	int nxt = -1;
	for(int i = 0; i < G[v].size(); ++i){
		int u = G[v][i];
		if(u == p) continue;
		if(sz[u] == 1) cnt++;
		else if(nxt == -1) nxt = u;
		else return false;
	}
	if(nxt == -1){
		for(int i = 1; i < cnt; ++i){
			perm.push_back(cur + i);
		}
		perm.push_back(cur);
		perm.push_back(cur + cnt);
		return true;
	}
	for(int i = 1; i <= cnt; ++i){
		perm.push_back(cur + i);
	}
	perm.push_back(cur);
	bool flg = dfs3(nxt, v, cur + cnt + 1);
	return flg;
}

vector<int> solve(){
	dfs1(0, -1);
	int d = -1, v = -1;
	for(int i = 0; i < N; ++i){
		if(d < dis[i]){
			d = dis[i];
			v = i;
		}
	}
	dis[v] = 0;
	dfs1(v, -1);
	d = -1;
	int u = -1;
	for(int i = 0; i < N; ++i){
		if(d < dis[i]){
			d = dis[i];
			u = i;
		}
	}
	
	vector<int> cand1;
	dfs(v, -1);
	bool flg = dfs3(v, -1, 1);
	if(flg) cand1 = perm;
	perm.clear();
	vector<int> cand2;
	dfs(u, -1);
	flg = dfs3(u, -1, 1);
	if(flg) cand2 = perm;
	
	return min(cand1, cand2);
}

int main(){
	input();
	vector<int> ans = solve();
	if(ans.size () == 0) printf("-1\n");
	else{
		for(int i = 0; i < N; ++i){
			printf("%d%c", ans[i], i == N - 1 ? '\n' : ' ');
		}
	}
	return 0;
}
