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
#include <bits/stdc++.h>
using namespace std;
#define M 100010
inline int read() {
	char ch = getchar(); int x = 0, f = 1;
	while(ch < '0' || ch > '9') {
		if(ch == '-') f = -1;
		ch = getchar();
	}
	while('0' <= ch && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x * f;
}
struct Edge{
	int u, v, Next;
} G[M * 2];
int head[M], tot;
inline void add(int u, int v) {
	G[++ tot] = (Edge){u, v, head[u]};
	head[u] = tot;
}
int n;
int Fa[M], d[M];
inline int dfs(int x, int fa) {
	Fa[x] = fa; d[x] = d[fa] + 1;
	int ret = x;
	for(int i = head[x]; i != -1; i = G[i].Next) {
		if(G[i].v == fa) continue;
		int v = dfs(G[i].v, x);
		if(d[v] > d[ret]) ret = v;
	}
	return ret;
}
bool is_[M]; 
int a[M], tt;
int cnt[M];
inline vector<int> Solve() {
	int now = 0;
	vector<int> ret;
	ret.resize(n);
	for(int i = 1; i <= tt; ++ i) {
		for(int j = 1; j < a[i]; ++ j) {
			ret[now + j - 1] = now + j + 1;
		}
		ret[now + a[i] - 1] = now + 1;
		now += a[i];
	} 
	return ret;
}
int main() {
//	freopen("a.in", "r", stdin);
	memset(head, -1, sizeof(head));
	n = read();
	for(int i = 1; i < n; ++ i) {
		int u = read(), v = read();
		add(u, v); add(v, u);
	}
	int s = dfs(1, 0);
	int t = dfs(s, 0);
//	cerr << s << " " << t << endl;
	for(int i = t; i != 0; i = Fa[i]) {
		is_[i] = 1;
	}
	for(int i = 1; i <= n; ++ i) {
		if(!is_[i] && !is_[Fa[i]]) {
			return puts("-1"), 0;
		}
		if(!is_[i]) cnt[Fa[i]] ++;
	}
	for(int i = t; i != 0; i = Fa[i]) {
		a[++ tt] = cnt[i] + 1;
	}
	vector<int> A1 = Solve();
	for(int i = 1; i <= tt / 2; ++ i) {
		swap(a[i], a[tt - i + 1]);
	}
	vector<int> A2 = Solve();
	for(int i = 0; i < A1.size(); ++ i) {
		if(A1[i] < A2[i]) {
			for(int j = 0; j < A1.size(); ++ j) {
				printf("%d ", A1[j]);
			}
			return 0;
		}
		if(A1[i] > A2[i]) {
			for(int j = 0; j < A2.size(); ++ j) {
				printf("%d ", A2[j]);
			}
			return 0;
		}
	}
	for(int j = 0; j < A1.size(); ++ j) {
		printf("%d ", A1[j]);
	}
}