#include <bits/stdc++.h>
#define eb emplace_back
#define allv(V) ((V).begin()),((V).end())
using namespace std;

const int MAXN = 100055;

vector<int> G[MAXN], AnsV;

int cnt[MAXN], prt[MAXN];

vector<int> Ans;
int N, rt1, rt2;

void dfs(int i) {
	for(int v : G[i]) if(!cnt[v]) {
		cnt[v] = cnt[i] + 1;
		dfs(v);
	}
}
void dfs1(int i) {
	cnt[i] = 1;
	for(int v : G[i]) if(!cnt[v]) {
		prt[v] = i;
		dfs1(v);
		cnt[i] += cnt[v];
	}
}

bool f(int i, int &c) {
	int t = 0;
	if(G[i].empty()) {
		AnsV.eb(++c);
		return true;
	}
	for(int v : G[i]) {
		if(!f(v, c)) return false;
		if(!t) AnsV.eb(cnt[i]);
		else if(1 != cnt[v]) return false;
		t++;
	}
	c++;
	return true;
}

void run(int rt) {
	memset(cnt, 0, sizeof cnt);
	AnsV.clear();

	dfs1(rt);
	for(int i = 1; i <= N; i++) {
		if(i != rt) G[i].erase(find(allv(G[i]), prt[i]));
		sort(allv(G[i]), [&](int a, int b) {
			return cnt[a] > cnt[b];
		});
	}
	{
		int c = 0;
		if(!f(rt, c)) return;
	}

	vector<int> V(N);
	for(int i = 0; i < N; i++)
		V[AnsV[i]-1] = i+1;
	
	if(Ans.empty() || V < Ans) swap(Ans, V);
}

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1, a, b; i < N; i++) {
		cin >> a >> b;
		G[a].eb(b);
		G[b].eb(a);
	}

	cnt[1] = 1; dfs(1);
	rt1 = int(max_element(cnt+1, cnt+N+1) - cnt);
	memset(cnt, 0, sizeof cnt);
	cnt[rt1] = 1; dfs(rt1);
	rt2 = int(max_element(cnt+1, cnt+N+1) - cnt);

	run(rt1);
	for(int i = 1; i <= N; i++) if(i != rt1) G[i].eb(prt[i]);
	run(rt2);

	if(Ans.empty()) puts("-1");
	else {
		for(int v : Ans) printf("%d ", v);
		puts("");
	}
	return 0;
}
### Prompt

Your task is to create 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 <bits/stdc++.h>
#define eb emplace_back
#define allv(V) ((V).begin()),((V).end())
using namespace std;

const int MAXN = 100055;

vector<int> G[MAXN], AnsV;

int cnt[MAXN], prt[MAXN];

vector<int> Ans;
int N, rt1, rt2;

void dfs(int i) {
	for(int v : G[i]) if(!cnt[v]) {
		cnt[v] = cnt[i] + 1;
		dfs(v);
	}
}
void dfs1(int i) {
	cnt[i] = 1;
	for(int v : G[i]) if(!cnt[v]) {
		prt[v] = i;
		dfs1(v);
		cnt[i] += cnt[v];
	}
}

bool f(int i, int &c) {
	int t = 0;
	if(G[i].empty()) {
		AnsV.eb(++c);
		return true;
	}
	for(int v : G[i]) {
		if(!f(v, c)) return false;
		if(!t) AnsV.eb(cnt[i]);
		else if(1 != cnt[v]) return false;
		t++;
	}
	c++;
	return true;
}

void run(int rt) {
	memset(cnt, 0, sizeof cnt);
	AnsV.clear();

	dfs1(rt);
	for(int i = 1; i <= N; i++) {
		if(i != rt) G[i].erase(find(allv(G[i]), prt[i]));
		sort(allv(G[i]), [&](int a, int b) {
			return cnt[a] > cnt[b];
		});
	}
	{
		int c = 0;
		if(!f(rt, c)) return;
	}

	vector<int> V(N);
	for(int i = 0; i < N; i++)
		V[AnsV[i]-1] = i+1;
	
	if(Ans.empty() || V < Ans) swap(Ans, V);
}

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1, a, b; i < N; i++) {
		cin >> a >> b;
		G[a].eb(b);
		G[b].eb(a);
	}

	cnt[1] = 1; dfs(1);
	rt1 = int(max_element(cnt+1, cnt+N+1) - cnt);
	memset(cnt, 0, sizeof cnt);
	cnt[rt1] = 1; dfs(rt1);
	rt2 = int(max_element(cnt+1, cnt+N+1) - cnt);

	run(rt1);
	for(int i = 1; i <= N; i++) if(i != rt1) G[i].eb(prt[i]);
	run(rt2);

	if(Ans.empty()) puts("-1");
	else {
		for(int v : Ans) printf("%d ", v);
		puts("");
	}
	return 0;
}
```