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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MN 101000

int cnt[MN], tot = 0;
int d[MN], d2[MN];
int h[MN], nxt[2 * MN], to[2 * MN], K = 0;
int ans1[MN], ans2[MN];

void ins(int u, int v) {nxt[++K] = h[u]; h[u] = K; to[K] = v;}

void dfs(int x, int fa)
{
	cnt[++tot] = d[x] - 2;
	for(int i = h[x]; i; i = nxt[i])
		if(to[i] != fa && d2[to[i]]) dfs(to[i], x);
}

void solve(int *ans)
{
	int now = 0;
	for(int i = 1; i <= tot; i++)
	{
		ans[now + cnt[i] + 1] = now + 1, now++;
		for(int j = 1; j <= cnt[i]; j++) ans[now] = now + 1, now++;
	}
}

int main()
{
	int n; scanf("%d", &n);
	if(n == 2) return 0 * puts("1 2"); 
	for(int i = 1; i < n; i++)
	{
		int u, v; scanf("%d%d", &u, &v);
		ins(u, v); ins(v, u);
		d[u]++; d[v]++;
	}
	memcpy(d2, d, sizeof(d));
	for(int i = 1; i <= n; i++)
		if(d[i] == 1) d2[i] = 0, d2[to[h[i]]]--;
	
	int id = 0;
	for(int i = 1; i <= n; i++)
		if(d2[i] > 2) return 0 * puts("-1");
		else if(d2[i] == 1) id = i;
	
	cnt[++tot] = 0; if(id) dfs(id, 0); else cnt[++tot] = n - 3; cnt[++tot] = 0;
	solve(ans1); std::reverse(cnt + 1, cnt + tot + 1); solve(ans2);

	bool ok = 0;
	for(int i = 1; i <= n; i++)
	{
		if(ans1[i] < ans2[i]) break;
		if(ans2[i] < ans1[i]) {ok = 1; break;}
	}

	if(ok) for(int i = 1; i <= n; i++) printf("%d ", ans2[i]);
	else for(int i = 1; i <= n; i++) printf("%d ", ans1[i]);
}