#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXN = 100000;

struct edge{
	int to; edge *nxt;
}edges[2*MAXN + 5], *adj[MAXN + 5], *ecnt = edges;
void addedge(int u, int v) {
	edge *p = (++ecnt);
	p->to = v, p->nxt = adj[u], adj[u] = p;
	p = (++ecnt);
	p->to = u, p->nxt = adj[v], adj[v] = p;
}

int deg[MAXN + 5], tmp[MAXN + 5];
int ans[MAXN + 5], a[MAXN + 5], cnt;
void dfs(int x, int f) {
	if( !f ) a[++cnt] = 1;
	int p = cnt + 1;
	for(int i=1;i<=deg[x]-2;i++)
		cnt++, a[cnt] = cnt + 1;
	a[++cnt] = p;
	if( f && tmp[x] == 1 ) cnt++, a[cnt] = cnt;
	for(edge *p=adj[x];p;p=p->nxt)
		if( deg[p->to] != 1 && p->to != f ) dfs(p->to, x);
}
void update() {
	for(int i=1;i<=cnt;i++) {
		if( ans[i] < a[i] ) break;
		else if( ans[i] > a[i] ) {
			for(int j=1;j<=cnt;j++)
				ans[j] = a[j];
			break;
		}
	}
	cnt = 0;
}

int main() {
	int n; scanf("%d", &n);
	for(int i=1;i<n;i++) {
		int u, v; scanf("%d%d", &u, &v);
		addedge(u, v), deg[u]++, deg[v]++;
	}
	
	if( n == 2 ) {
		printf("%d %d\n", 1, 2);
		return 0;
	}
	for(int i=1;i<=n;i++) tmp[i] = deg[i];
	for(int i=1;i<=n;i++)
		if( deg[i] == 1 ) tmp[adj[i]->to]--;
	for(int i=1;i<=n;i++)
		if( deg[i] != 1 && tmp[i] > 2 ) {
			puts("-1");
			return 0;
		}
	
	int s = 0, t = 0;
	for(int i=1;i<=n;i++)
		if( deg[i] != 1 && tmp[i] == 1 ) {
			if( !s ) s = i;
			else t = i;
		}
	if( t == 0 ) {
		printf("1");
		for(int i=3;i<n;i++)
			printf(" %d", i);
		printf(" %d %d\n", 2, n);
	}
	else {
		for(int i=1;i<=n;i++) ans[i] = n - i + 1;
		dfs(s, 0), update(), dfs(t, 0), update();
		for(int i=1;i<=n;i++)
			printf("%d%c", ans[i], i == n ? '\n' : ' ');
	}
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <algorithm>
using namespace std;

const int MAXN = 100000;

struct edge{
	int to; edge *nxt;
}edges[2*MAXN + 5], *adj[MAXN + 5], *ecnt = edges;
void addedge(int u, int v) {
	edge *p = (++ecnt);
	p->to = v, p->nxt = adj[u], adj[u] = p;
	p = (++ecnt);
	p->to = u, p->nxt = adj[v], adj[v] = p;
}

int deg[MAXN + 5], tmp[MAXN + 5];
int ans[MAXN + 5], a[MAXN + 5], cnt;
void dfs(int x, int f) {
	if( !f ) a[++cnt] = 1;
	int p = cnt + 1;
	for(int i=1;i<=deg[x]-2;i++)
		cnt++, a[cnt] = cnt + 1;
	a[++cnt] = p;
	if( f && tmp[x] == 1 ) cnt++, a[cnt] = cnt;
	for(edge *p=adj[x];p;p=p->nxt)
		if( deg[p->to] != 1 && p->to != f ) dfs(p->to, x);
}
void update() {
	for(int i=1;i<=cnt;i++) {
		if( ans[i] < a[i] ) break;
		else if( ans[i] > a[i] ) {
			for(int j=1;j<=cnt;j++)
				ans[j] = a[j];
			break;
		}
	}
	cnt = 0;
}

int main() {
	int n; scanf("%d", &n);
	for(int i=1;i<n;i++) {
		int u, v; scanf("%d%d", &u, &v);
		addedge(u, v), deg[u]++, deg[v]++;
	}
	
	if( n == 2 ) {
		printf("%d %d\n", 1, 2);
		return 0;
	}
	for(int i=1;i<=n;i++) tmp[i] = deg[i];
	for(int i=1;i<=n;i++)
		if( deg[i] == 1 ) tmp[adj[i]->to]--;
	for(int i=1;i<=n;i++)
		if( deg[i] != 1 && tmp[i] > 2 ) {
			puts("-1");
			return 0;
		}
	
	int s = 0, t = 0;
	for(int i=1;i<=n;i++)
		if( deg[i] != 1 && tmp[i] == 1 ) {
			if( !s ) s = i;
			else t = i;
		}
	if( t == 0 ) {
		printf("1");
		for(int i=3;i<n;i++)
			printf(" %d", i);
		printf(" %d %d\n", 2, n);
	}
	else {
		for(int i=1;i<=n;i++) ans[i] = n - i + 1;
		dfs(s, 0), update(), dfs(t, 0), update();
		for(int i=1;i<=n;i++)
			printf("%d%c", ans[i], i == n ? '\n' : ' ');
	}
}
```