#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring> 
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 200010;
ll read() {
	ll x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1; c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0'; c = getchar();
	}
	return x * f;
}
struct node {
	int v,nxt;
}e[N << 1];
int du[N],ejs,head[N];
void add(int u,int v) {
	e[++ejs].v = v;e[ejs].nxt = head[u];head[u] = ejs;
}
int n,dis[N];
void dfs(int u,int fa) {
	for(int i = head[u];i;i = e[i].nxt) {
		int v = e[i].v;if(v == fa) continue;
		dis[v] = dis[u] + 1;
		dfs(v,u);
	}
}
int zj[N],tot;
bool get(int u,int fa,int T) {
	if(u == T) {
		zj[++tot] = T;
		return true;
	}
	for(int i = head[u];i;i = e[i].nxt) {
		int v = e[i].v;if(v == fa) continue;
		if(get(v,u,T)) {
			zj[++tot] = u;
			return true;
		}
	}
	return false;
}
int ans[N];
int main() {
	n = read();
	for(int i = 1;i < n;++i) {
		int u = read(),v = read();add(v,u);add(u,v);
		du[u]++;du[v]++;
	}
	
	dfs(1,0);
	int tmp = 0;
	for(int i = 1;i <= n;++i)
		if(dis[i] > dis[tmp]) tmp = i;
	dis[tmp] = 0;
	dfs(tmp,0);
	int tt = 0;
	for(int i = 1;i <= n;++i) 
		if(dis[i] > dis[tt]) tt = i;
	get(tmp,0,tt);
	
	int flag = 0;
	for(int i = 1;i <= tot;++i) {
		if(du[zj[i]] < du[zj[tot - i + 1]]) break;
		if(du[zj[i]] > du[zj[tot - i + 1]]) {
			flag = 1;break;
		}
	}
	if(flag) {
		for(int i = 1;i <= tot / 2;++i) swap(zj[i],zj[tot - i + 1]);
	}
	
	int pos = 0,now = 0;
	
	ans[++pos] = ++now;
	for(int i = 2;i < tot;++i) {
		int tmp = ++now;
		for(int j = 1;j < du[zj[i]] - 1;++j) ans[++pos] = ++now;
		ans[++pos] = tmp;
	}
	ans[++pos] = ++now;
	if(pos != n) {
		puts("-1");return 0;
	}
	for(int i = 1;i <= pos;++i) printf("%d ",ans[i]);
	return 0;
}

### Prompt

Please formulate a Cpp solution to 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<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring> 
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 200010;
ll read() {
	ll x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1; c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0'; c = getchar();
	}
	return x * f;
}
struct node {
	int v,nxt;
}e[N << 1];
int du[N],ejs,head[N];
void add(int u,int v) {
	e[++ejs].v = v;e[ejs].nxt = head[u];head[u] = ejs;
}
int n,dis[N];
void dfs(int u,int fa) {
	for(int i = head[u];i;i = e[i].nxt) {
		int v = e[i].v;if(v == fa) continue;
		dis[v] = dis[u] + 1;
		dfs(v,u);
	}
}
int zj[N],tot;
bool get(int u,int fa,int T) {
	if(u == T) {
		zj[++tot] = T;
		return true;
	}
	for(int i = head[u];i;i = e[i].nxt) {
		int v = e[i].v;if(v == fa) continue;
		if(get(v,u,T)) {
			zj[++tot] = u;
			return true;
		}
	}
	return false;
}
int ans[N];
int main() {
	n = read();
	for(int i = 1;i < n;++i) {
		int u = read(),v = read();add(v,u);add(u,v);
		du[u]++;du[v]++;
	}
	
	dfs(1,0);
	int tmp = 0;
	for(int i = 1;i <= n;++i)
		if(dis[i] > dis[tmp]) tmp = i;
	dis[tmp] = 0;
	dfs(tmp,0);
	int tt = 0;
	for(int i = 1;i <= n;++i) 
		if(dis[i] > dis[tt]) tt = i;
	get(tmp,0,tt);
	
	int flag = 0;
	for(int i = 1;i <= tot;++i) {
		if(du[zj[i]] < du[zj[tot - i + 1]]) break;
		if(du[zj[i]] > du[zj[tot - i + 1]]) {
			flag = 1;break;
		}
	}
	if(flag) {
		for(int i = 1;i <= tot / 2;++i) swap(zj[i],zj[tot - i + 1]);
	}
	
	int pos = 0,now = 0;
	
	ans[++pos] = ++now;
	for(int i = 2;i < tot;++i) {
		int tmp = ++now;
		for(int j = 1;j < du[zj[i]] - 1;++j) ans[++pos] = ++now;
		ans[++pos] = tmp;
	}
	ans[++pos] = ++now;
	if(pos != n) {
		puts("-1");return 0;
	}
	for(int i = 1;i <= pos;++i) printf("%d ",ans[i]);
	return 0;
}

```