#include<bits/stdc++.h>
using namespace std;
#define ran 100010
int n,U,W;
vector<int> e[ran];
int d[ran],fat[ran];
bool in[ran];int mp[ran],sz[ran],lg;
void dfs(int x){
	if(d[x] > d[W])W = x;
	for(auto y : e[x]){
		if(y == fat[x])continue;
		d[y] = d[x] + 1;
		fat[y] = x;
		dfs(y);
	}
}
int A[ran],B[ran];
void doit(int*A){
	int w = 0,l = 0;
	for(int i=0; i<lg; i++){
		for(int j=0; j<sz[i]; j++)
			A[l++] = w + 2 + j;
		A[l++] = w + 1;
		w += 1 + sz[i];
	}
}
bool chk(){
	lg = 0;
	while(1){
		in[W] = true;
		mp[W] = lg++;
		if(W == U)break;
		W = fat[W];
	}
	for(int i=1; i<=n; i++){
		if(!in[i]){
			if(!in[fat[i]])
				return false;
			sz[mp[fat[i]]] ++;
		}
	}
	doit(A);
	reverse(sz,sz+lg);
	doit(B);
	if(memcmp(A,B,sizeof(int)*n) > 0)
		memcpy(A,B,sizeof(int)*n);
	for(int i=0; i<n; i++)
		printf("%d%c",A[i],i<n-1?' ':'\n');
	return true;
}
int main(){
	scanf("%d",&n);
	for(int i=1; i<n; i++){
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	fat[1] = W = 1;
	d[1] = 0;
	dfs(1);
	U = W;
	fat[U] = U; d[U] = 0;
	dfs(U);
	if(!chk()){
		puts("-1");
		return 0;
	}
	return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define ran 100010
int n,U,W;
vector<int> e[ran];
int d[ran],fat[ran];
bool in[ran];int mp[ran],sz[ran],lg;
void dfs(int x){
	if(d[x] > d[W])W = x;
	for(auto y : e[x]){
		if(y == fat[x])continue;
		d[y] = d[x] + 1;
		fat[y] = x;
		dfs(y);
	}
}
int A[ran],B[ran];
void doit(int*A){
	int w = 0,l = 0;
	for(int i=0; i<lg; i++){
		for(int j=0; j<sz[i]; j++)
			A[l++] = w + 2 + j;
		A[l++] = w + 1;
		w += 1 + sz[i];
	}
}
bool chk(){
	lg = 0;
	while(1){
		in[W] = true;
		mp[W] = lg++;
		if(W == U)break;
		W = fat[W];
	}
	for(int i=1; i<=n; i++){
		if(!in[i]){
			if(!in[fat[i]])
				return false;
			sz[mp[fat[i]]] ++;
		}
	}
	doit(A);
	reverse(sz,sz+lg);
	doit(B);
	if(memcmp(A,B,sizeof(int)*n) > 0)
		memcpy(A,B,sizeof(int)*n);
	for(int i=0; i<n; i++)
		printf("%d%c",A[i],i<n-1?' ':'\n');
	return true;
}
int main(){
	scanf("%d",&n);
	for(int i=1; i<n; i++){
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	fat[1] = W = 1;
	d[1] = 0;
	dfs(1);
	U = W;
	fat[U] = U; d[U] = 0;
	dfs(U);
	if(!chk()){
		puts("-1");
		return 0;
	}
	return 0;
}

```