#include <cstdio>
#include <algorithm>
#include <vector>

const int MN = 100005;

int N;
std::vector<int> G[MN];
int dep[MN], faz[MN], o[MN], cnt[MN], tp;

void DFS(int u, int fz) {
	dep[u] = dep[faz[u] = fz] + 1;
	for (auto v : G[u]) if (v != fz) DFS(v, u);
}

int main() {
	scanf("%d", &N);
	for (int i = 1, x, y; i < N; ++i)
		scanf("%d%d", &x, &y),
		G[x].push_back(y),
		G[y].push_back(x);
	DFS(1, 0);
	int u = 0, v = 0;
	for (int i = 1; i <= N; ++i) if (dep[u] < dep[i]) u = i;
	DFS(u, 0);
	for (int i = 1; i <= N; ++i) if (dep[v] < dep[i]) v = i;
	for (int x = v; x != u; x = faz[x]) o[x] = ++tp;
	o[u] = ++tp;
	for (int i = 1; i <= N; ++i) if (!o[i]) {
		if (G[i].size() == 1) ++cnt[o[G[i][0]]];
		else return puts("-1"), 0;
	}
	int rev = 0;
	for (int i = 1, j = tp; i < j; ++i, --j)
		if (cnt[i] != cnt[j]) { rev = cnt[i] > cnt[j]; break; }
	if (rev) std::reverse(cnt + 1, cnt + tp + 1);
	int now = 1;
	for (int i = 1; i <= tp; ++i) {
		for (int j = 1; j <= cnt[i]; ++j) printf("%d ", now + j);
		printf("%d ", now), now += cnt[i] + 1;
	}
	return 0;
}
### Prompt

In CPP, your task is to solve 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 <cstdio>
#include <algorithm>
#include <vector>

const int MN = 100005;

int N;
std::vector<int> G[MN];
int dep[MN], faz[MN], o[MN], cnt[MN], tp;

void DFS(int u, int fz) {
	dep[u] = dep[faz[u] = fz] + 1;
	for (auto v : G[u]) if (v != fz) DFS(v, u);
}

int main() {
	scanf("%d", &N);
	for (int i = 1, x, y; i < N; ++i)
		scanf("%d%d", &x, &y),
		G[x].push_back(y),
		G[y].push_back(x);
	DFS(1, 0);
	int u = 0, v = 0;
	for (int i = 1; i <= N; ++i) if (dep[u] < dep[i]) u = i;
	DFS(u, 0);
	for (int i = 1; i <= N; ++i) if (dep[v] < dep[i]) v = i;
	for (int x = v; x != u; x = faz[x]) o[x] = ++tp;
	o[u] = ++tp;
	for (int i = 1; i <= N; ++i) if (!o[i]) {
		if (G[i].size() == 1) ++cnt[o[G[i][0]]];
		else return puts("-1"), 0;
	}
	int rev = 0;
	for (int i = 1, j = tp; i < j; ++i, --j)
		if (cnt[i] != cnt[j]) { rev = cnt[i] > cnt[j]; break; }
	if (rev) std::reverse(cnt + 1, cnt + tp + 1);
	int now = 1;
	for (int i = 1; i <= tp; ++i) {
		for (int j = 1; j <= cnt[i]; ++j) printf("%d ", now + j);
		printf("%d ", now), now += cnt[i] + 1;
	}
	return 0;
}
```