#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int n, a[100009], b[100009], G[100009], ans[100009];
vector<int>x[100009], y[100009]; bool used[100009]; vector<int>V;

vector<int>solve(vector<int>Z) {
	vector<int>J; J.push_back(0);
	for (int i = 0; i < Z.size(); i++) Z[i] -= 2;
	for (int i = 0; i < Z.size(); i++) {
		int B = J.size();
		for (int j = 0; j < Z[i]; j++) J.push_back(B + j + 1);
		J.push_back(B);
	}
	J.push_back(n - 1);
	return J;
}

int main() {
	cin >> n;
	for (int i = 1; i <= n - 1; i++) {
		cin >> a[i] >> b[i];
		G[a[i]]++; G[b[i]]++;
		y[a[i]].push_back(b[i]);
		y[b[i]].push_back(a[i]);
	}
	int t = 0, eg = 0;
	for (int i = 1; i <= n - 1; i++) {
		if (G[a[i]] >= 2 && G[b[i]] >= 2) {
			x[a[i]].push_back(b[i]);
			x[b[i]].push_back(a[i]);
			eg++;
		}
	}
	int ver = 0;
	for (int i = 1; i <= n; i++) {
		if (x[i].size() >= 1) ver++;
		if (y[i].size() >= 2) t = i;
	}
	if (!(ver == eg + 1 || ver == 0)) {
		cout << "-1" << endl;
		return 0;
	}
	int s = 0;
	for (int i = 1; i <= n; i++) {
		if (x[i].size() >= 3) {
			cout << "-1" << endl;
			return 0;
		}
		if (x[i].size() == 1) s = i;
	}
	if (s == 0) s = t;
	used[s] = true; V.push_back(y[s].size());

	while (true) {
		bool flag = false;
		for (int i = 0; i < x[s].size(); i++) {
			if (used[x[s][i]] == false) {
				used[x[s][i]] = true;
				s = x[s][i]; flag = true;
				break;
			}
		}
		if (flag == true) V.push_back(y[s].size());
		else break;
	}
	vector<int>F1 = solve(V); reverse(V.begin(), V.end());
	vector<int>F2 = solve(V);
	vector<int>F = min(F1, F2);
	for (int i = 0; i < F.size(); i++) { if (i)cout << " "; cout << F[i] + 1; }cout << endl;
	return 0;
}
### Prompt

Please create a solution in Cpp 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int n, a[100009], b[100009], G[100009], ans[100009];
vector<int>x[100009], y[100009]; bool used[100009]; vector<int>V;

vector<int>solve(vector<int>Z) {
	vector<int>J; J.push_back(0);
	for (int i = 0; i < Z.size(); i++) Z[i] -= 2;
	for (int i = 0; i < Z.size(); i++) {
		int B = J.size();
		for (int j = 0; j < Z[i]; j++) J.push_back(B + j + 1);
		J.push_back(B);
	}
	J.push_back(n - 1);
	return J;
}

int main() {
	cin >> n;
	for (int i = 1; i <= n - 1; i++) {
		cin >> a[i] >> b[i];
		G[a[i]]++; G[b[i]]++;
		y[a[i]].push_back(b[i]);
		y[b[i]].push_back(a[i]);
	}
	int t = 0, eg = 0;
	for (int i = 1; i <= n - 1; i++) {
		if (G[a[i]] >= 2 && G[b[i]] >= 2) {
			x[a[i]].push_back(b[i]);
			x[b[i]].push_back(a[i]);
			eg++;
		}
	}
	int ver = 0;
	for (int i = 1; i <= n; i++) {
		if (x[i].size() >= 1) ver++;
		if (y[i].size() >= 2) t = i;
	}
	if (!(ver == eg + 1 || ver == 0)) {
		cout << "-1" << endl;
		return 0;
	}
	int s = 0;
	for (int i = 1; i <= n; i++) {
		if (x[i].size() >= 3) {
			cout << "-1" << endl;
			return 0;
		}
		if (x[i].size() == 1) s = i;
	}
	if (s == 0) s = t;
	used[s] = true; V.push_back(y[s].size());

	while (true) {
		bool flag = false;
		for (int i = 0; i < x[s].size(); i++) {
			if (used[x[s][i]] == false) {
				used[x[s][i]] = true;
				s = x[s][i]; flag = true;
				break;
			}
		}
		if (flag == true) V.push_back(y[s].size());
		else break;
	}
	vector<int>F1 = solve(V); reverse(V.begin(), V.end());
	vector<int>F2 = solve(V);
	vector<int>F = min(F1, F2);
	for (int i = 0; i < F.size(); i++) { if (i)cout << " "; cout << F[i] + 1; }cout << endl;
	return 0;
}
```