#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/
const int MAXN = 120000;

vector<int> eds[MAXN];
int n;

pair<int, int> dfs1(int v, int p) {
	pair<int, int> ans = {0, v};
	for (int u: eds[v]) {
		if (u == p)
			continue;
		pair<int, int> tmp = dfs1(u, v);
		++tmp.first;
		ans = max(ans, tmp);
	}
	return ans;
}
vector<int> st;

int get(int a, int p, int b) {
	st.push_back(a);
	if (a == b)
		return 1;
	for (int u: eds[a]) {
		if (u == p)
			continue;
		if (get(u, a, b))
			return 1;
	}
	st.pop_back();
	return 0;
}

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n;
	for (int i = 1; i < n; ++i) {
		int a, b;
		cin >> a >> b;
		--a, --b;
		eds[a].push_back(b);
		eds[b].push_back(a);
	}
	int a = dfs1(0, -1).second;
	int b = dfs1(a, -1).second;
	st.clear();
	get(a, -1, b);
	vector<int> vv;
	int sum = st.size();
	for (int i = 1; i < st.size() - 1; ++i) {
		sum += eds[st[i]].size() - 2;
		vv.push_back(eds[st[i]].size() - 2);
	}
	if (sum != n) {
		cout << -1 << "\n";
		return 0;
	}
	vector<int> vv2 = vv;
	reverse(ALL(vv2));
	if (vv > vv2)
		vv = vv2, reverse(ALL(st));
	vv.insert(vv.begin(), 0);
	vv.push_back(0);
	int cur = 0;
	vector<int> ans;
	for (int i = 0; i < st.size(); ++i) {
		int start = cur;
		++cur;
		for (int j = 0; j < vv[i]; ++j)
			ans.push_back(cur++);
		ans.push_back(start);
	}
	for (int i: ans)
		cout << i + 1 << " ";
	cout << "\n";
	return 0;
}



### Prompt

Create a solution in CPP for 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
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/
const int MAXN = 120000;

vector<int> eds[MAXN];
int n;

pair<int, int> dfs1(int v, int p) {
	pair<int, int> ans = {0, v};
	for (int u: eds[v]) {
		if (u == p)
			continue;
		pair<int, int> tmp = dfs1(u, v);
		++tmp.first;
		ans = max(ans, tmp);
	}
	return ans;
}
vector<int> st;

int get(int a, int p, int b) {
	st.push_back(a);
	if (a == b)
		return 1;
	for (int u: eds[a]) {
		if (u == p)
			continue;
		if (get(u, a, b))
			return 1;
	}
	st.pop_back();
	return 0;
}

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n;
	for (int i = 1; i < n; ++i) {
		int a, b;
		cin >> a >> b;
		--a, --b;
		eds[a].push_back(b);
		eds[b].push_back(a);
	}
	int a = dfs1(0, -1).second;
	int b = dfs1(a, -1).second;
	st.clear();
	get(a, -1, b);
	vector<int> vv;
	int sum = st.size();
	for (int i = 1; i < st.size() - 1; ++i) {
		sum += eds[st[i]].size() - 2;
		vv.push_back(eds[st[i]].size() - 2);
	}
	if (sum != n) {
		cout << -1 << "\n";
		return 0;
	}
	vector<int> vv2 = vv;
	reverse(ALL(vv2));
	if (vv > vv2)
		vv = vv2, reverse(ALL(st));
	vv.insert(vv.begin(), 0);
	vv.push_back(0);
	int cur = 0;
	vector<int> ans;
	for (int i = 0; i < st.size(); ++i) {
		int start = cur;
		++cur;
		for (int j = 0; j < vv[i]; ++j)
			ans.push_back(cur++);
		ans.push_back(start);
	}
	for (int i: ans)
		cout << i + 1 << " ";
	cout << "\n";
	return 0;
}



```