#include <bits/stdc++.h>

const int MAXN = 100010;
void bye() { std::cout << -1 << std::endl; exit(0); }
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
int deg[MAXN], n;
void addedge(int b, int e) {
	nxt[++tot] = head[b]; to[head[b] = tot] = e;
	nxt[++tot] = head[e]; to[head[e] = tot] = b;
	++deg[b]; ++deg[e];
}
typedef std::pair<int, int> pi;
int son[MAXN], fa[MAXN], ind[MAXN], li[MAXN], bak;
pi diameter(int u) {
	pi res(0, u); son[u] = 0;
	for (int i = head[u]; i; i = nxt[i])
		if (to[i] != fa[u]) {
			fa[to[i]] = u;
			pi ret = diameter(to[i]); ++ret.first;
			if (ret > res) {
				son[u] = to[i];
				res = ret;
			}
		}
	return res;
}
int ansl[MAXN], tmp[MAXN];
void solve() {
	int now = 0, tl = 0;
	for (int i = 1; i <= bak; ++i) {
		for (int j = now + 2; j <= now + deg[li[i]] - (i != 1); ++j)
			tmp[++tl] = j;
		tmp[++tl] = now + 1;
		now += deg[li[i]] - (i != 1);
	}
	if (std::lexicographical_compare(tmp + 1, tmp + 1 + n, ansl + 1, ansl + 1 + n))
		memcpy(ansl, tmp, n + 1 << 2);
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1, t1, t2; i < n; ++i)
		std::cin >> t1 >> t2, addedge(t1, t2);
	int rtl = diameter(1).second;
	memset(fa, 0, n + 1 << 2);
	diameter(rtl).second;
	ind[li[++bak] = rtl] = true; int now = rtl;
	ind[0] = true;
	while (son[now]) ind[li[++bak] = son[now]] = true, now = son[now];
	for (int i = 1; i <= n; ++i) if (!ind[fa[i]]) bye();
	ansl[1] = 123456789;
	solve(); std::reverse(li + 1, li + 1 + bak); solve();
	for (int i = 1; i <= n; ++i) std::cout << ansl[i] << ' ';
	std::cout << std::endl;
	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 <bits/stdc++.h>

const int MAXN = 100010;
void bye() { std::cout << -1 << std::endl; exit(0); }
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
int deg[MAXN], n;
void addedge(int b, int e) {
	nxt[++tot] = head[b]; to[head[b] = tot] = e;
	nxt[++tot] = head[e]; to[head[e] = tot] = b;
	++deg[b]; ++deg[e];
}
typedef std::pair<int, int> pi;
int son[MAXN], fa[MAXN], ind[MAXN], li[MAXN], bak;
pi diameter(int u) {
	pi res(0, u); son[u] = 0;
	for (int i = head[u]; i; i = nxt[i])
		if (to[i] != fa[u]) {
			fa[to[i]] = u;
			pi ret = diameter(to[i]); ++ret.first;
			if (ret > res) {
				son[u] = to[i];
				res = ret;
			}
		}
	return res;
}
int ansl[MAXN], tmp[MAXN];
void solve() {
	int now = 0, tl = 0;
	for (int i = 1; i <= bak; ++i) {
		for (int j = now + 2; j <= now + deg[li[i]] - (i != 1); ++j)
			tmp[++tl] = j;
		tmp[++tl] = now + 1;
		now += deg[li[i]] - (i != 1);
	}
	if (std::lexicographical_compare(tmp + 1, tmp + 1 + n, ansl + 1, ansl + 1 + n))
		memcpy(ansl, tmp, n + 1 << 2);
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1, t1, t2; i < n; ++i)
		std::cin >> t1 >> t2, addedge(t1, t2);
	int rtl = diameter(1).second;
	memset(fa, 0, n + 1 << 2);
	diameter(rtl).second;
	ind[li[++bak] = rtl] = true; int now = rtl;
	ind[0] = true;
	while (son[now]) ind[li[++bak] = son[now]] = true, now = son[now];
	for (int i = 1; i <= n; ++i) if (!ind[fa[i]]) bye();
	ansl[1] = 123456789;
	solve(); std::reverse(li + 1, li + 1 + bak); solve();
	for (int i = 1; i <= n; ++i) std::cout << ansl[i] << ' ';
	std::cout << std::endl;
	return 0;
}
```