#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#define PB push_back
#define MP make_pair
#define FIR first
#define SEC second
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=1e5+10;
int n;
vector<int> G[N];
int mxd,cur;
void dfs1(int u,int last,int d) {
	if(d>mxd) mxd=d,cur=u;
	for(int i=0;i<G[u].size();++i)
		if(G[u][i]!=last) dfs1(G[u][i],u,d+1);
}
int st[N],top;
int mk[N];
vector<int> L;
void dfs2(int u,int last,int T) {
	st[++top]=u;
	if(u==T) {
		for(int i=1;i<=top;++i) mk[st[i]]=1,L.PB(st[i]);
	}
	for(int i=0;i<G[u].size();++i)
		if(G[u][i]!=last) dfs2(G[u][i],u,T);
	top--;
}
void sol(int *P) {
	int m=0;
	P[++m]=1;
	for(int i=1;i+1<L.size();++i) {
		int l=m+1,r=m+G[L[i]].size()-1;
		P[++m]=r;
		for(int j=l;j<r;++j) P[++m]=j;
	}
	P[++m]=n;
}
int P[N],Q[N];
int main() {
	rd(n);
	for(int i=1,x,y;i<n;++i) rd(x),rd(y),G[x].PB(y),G[y].PB(x);
	mxd=-1; dfs1(1,0,0); int L1=cur;
	mxd=-1; dfs1(cur,0,0); int L2=cur;
	dfs2(L1,0,L2);
	for(int i=1;i<=n;++i) {
		int flg=mk[i];
		for(int j=0;!flg&&j<G[i].size();++j)
			flg|=mk[G[i][j]];
		if(!flg) {
			printf("-1");
			return 0;
		}
	}
	sol(P);
	reverse(L.begin(),L.end());
	sol(Q);
	for(int i=1;i<=n;++i)
		if(P[i]!=Q[i]) {
			if(P[i]>Q[i]) swap(P,Q);
			break;
		}
	for(int i=1;i<=n;++i) Q[P[i]]=i;
	for(int i=1;i<=n;++i) printf("%d ",Q[i]);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#define PB push_back
#define MP make_pair
#define FIR first
#define SEC second
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=1e5+10;
int n;
vector<int> G[N];
int mxd,cur;
void dfs1(int u,int last,int d) {
	if(d>mxd) mxd=d,cur=u;
	for(int i=0;i<G[u].size();++i)
		if(G[u][i]!=last) dfs1(G[u][i],u,d+1);
}
int st[N],top;
int mk[N];
vector<int> L;
void dfs2(int u,int last,int T) {
	st[++top]=u;
	if(u==T) {
		for(int i=1;i<=top;++i) mk[st[i]]=1,L.PB(st[i]);
	}
	for(int i=0;i<G[u].size();++i)
		if(G[u][i]!=last) dfs2(G[u][i],u,T);
	top--;
}
void sol(int *P) {
	int m=0;
	P[++m]=1;
	for(int i=1;i+1<L.size();++i) {
		int l=m+1,r=m+G[L[i]].size()-1;
		P[++m]=r;
		for(int j=l;j<r;++j) P[++m]=j;
	}
	P[++m]=n;
}
int P[N],Q[N];
int main() {
	rd(n);
	for(int i=1,x,y;i<n;++i) rd(x),rd(y),G[x].PB(y),G[y].PB(x);
	mxd=-1; dfs1(1,0,0); int L1=cur;
	mxd=-1; dfs1(cur,0,0); int L2=cur;
	dfs2(L1,0,L2);
	for(int i=1;i<=n;++i) {
		int flg=mk[i];
		for(int j=0;!flg&&j<G[i].size();++j)
			flg|=mk[G[i][j]];
		if(!flg) {
			printf("-1");
			return 0;
		}
	}
	sol(P);
	reverse(L.begin(),L.end());
	sol(Q);
	for(int i=1;i<=n;++i)
		if(P[i]!=Q[i]) {
			if(P[i]>Q[i]) swap(P,Q);
			break;
		}
	for(int i=1;i<=n;++i) Q[P[i]]=i;
	for(int i=1;i<=n;++i) printf("%d ",Q[i]);
	return 0;
}
```