#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define E "\n"

using namespace std;
const long long MOD = (long long)1e9 + 7;

int n, u, v;
vector<int> vec[100009], ans[2], sp;
bool mk[100009];

bool ck(int x, bool fst)
{
	int cnt = 0;
	mk[x] = true;
	for (int i = 0; i < (int)vec[x].size(); i++) {
		if (!mk[vec[x][i]]) {
			if (!ck(vec[x][i], false)) return false;
			cnt++;
			if (cnt > 1 && !fst) return false;
			if (cnt > 2 && fst) return false;
		}
	}
	if (cnt == 0) sp.pb(x);
	if (cnt == 1 && fst) sp.pb(x);
	return true;
}

void fun(int cur, int cur2, int id)
{
	int cnt = vec[cur].size() - 1;
	if (cur2 < n) cnt--;
	int tmp = cur2 + 1;
	mk[cur] = true;
	while (cnt) {
		ans[id].pb(tmp);
		tmp++;
		cnt--;
	}
	ans[id].pb(cur2);
	if (tmp <= n) {
		for (int i = 0; i < (int)vec[cur].size(); i++) {
			if (!mk[vec[cur][i]]) {
				fun(vec[cur][i], tmp, id);
				break;
			}
		}
	}
}

int main()
{
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		cin >> u >> v;
		vec[u].pb(v);
		vec[v].pb(u);
	}
	if (n == 2) {
		cout << "1 2" << E;
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		if (vec[i].size() == 1) mk[i] = true;
	}
	bool ok = true;
	for (int i = 1; i <= n; i++) {
		if (!mk[i]) {
			ok = ck(i, true);
			break;
		}
	}
	if (!ok) {
		cout << -1 << E;
	}
	else {
		if (sp.size() == 1) sp.pb(sp[0]);
		memset(mk, 0, sizeof(mk));
		for (int i = 1; i <= n; i++) {
			if (vec[i].size() == 1) mk[i] = true;
		}
		ans[0].pb(1);
		fun(sp[0], 2, 0);
		ans[0].pb(n);
		memset(mk, 0, sizeof(mk));
		for (int i = 1; i <= n; i++) {
			if (vec[i].size() == 1) mk[i] = true;
		}
		ans[1].pb(1);
		fun(sp[1], 2, 1);
		ans[1].pb(n);
		int tmp = 0;
		for (int i = 0; i < n; i++) {
			if (ans[1][i] == ans[0][i]) continue;
			if (ans[1][i] < ans[0][i]) {
				tmp = 1;
				break;
			}
			else break;
		}
		for (int i = 0; i < n; i++) cout << ans[tmp][i] << " ";
	}
	//system("pause");
	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
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define E "\n"

using namespace std;
const long long MOD = (long long)1e9 + 7;

int n, u, v;
vector<int> vec[100009], ans[2], sp;
bool mk[100009];

bool ck(int x, bool fst)
{
	int cnt = 0;
	mk[x] = true;
	for (int i = 0; i < (int)vec[x].size(); i++) {
		if (!mk[vec[x][i]]) {
			if (!ck(vec[x][i], false)) return false;
			cnt++;
			if (cnt > 1 && !fst) return false;
			if (cnt > 2 && fst) return false;
		}
	}
	if (cnt == 0) sp.pb(x);
	if (cnt == 1 && fst) sp.pb(x);
	return true;
}

void fun(int cur, int cur2, int id)
{
	int cnt = vec[cur].size() - 1;
	if (cur2 < n) cnt--;
	int tmp = cur2 + 1;
	mk[cur] = true;
	while (cnt) {
		ans[id].pb(tmp);
		tmp++;
		cnt--;
	}
	ans[id].pb(cur2);
	if (tmp <= n) {
		for (int i = 0; i < (int)vec[cur].size(); i++) {
			if (!mk[vec[cur][i]]) {
				fun(vec[cur][i], tmp, id);
				break;
			}
		}
	}
}

int main()
{
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		cin >> u >> v;
		vec[u].pb(v);
		vec[v].pb(u);
	}
	if (n == 2) {
		cout << "1 2" << E;
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		if (vec[i].size() == 1) mk[i] = true;
	}
	bool ok = true;
	for (int i = 1; i <= n; i++) {
		if (!mk[i]) {
			ok = ck(i, true);
			break;
		}
	}
	if (!ok) {
		cout << -1 << E;
	}
	else {
		if (sp.size() == 1) sp.pb(sp[0]);
		memset(mk, 0, sizeof(mk));
		for (int i = 1; i <= n; i++) {
			if (vec[i].size() == 1) mk[i] = true;
		}
		ans[0].pb(1);
		fun(sp[0], 2, 0);
		ans[0].pb(n);
		memset(mk, 0, sizeof(mk));
		for (int i = 1; i <= n; i++) {
			if (vec[i].size() == 1) mk[i] = true;
		}
		ans[1].pb(1);
		fun(sp[1], 2, 1);
		ans[1].pb(n);
		int tmp = 0;
		for (int i = 0; i < n; i++) {
			if (ans[1][i] == ans[0][i]) continue;
			if (ans[1][i] < ans[0][i]) {
				tmp = 1;
				break;
			}
			else break;
		}
		for (int i = 0; i < n; i++) cout << ans[tmp][i] << " ";
	}
	//system("pause");
	return 0;
}

```