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>

const int N = 100054, M = N * 2;

int n, E = 0;
int to[M], first[N], next[M];
int p[N], dep[N], deg[N];
int cnt = 0, a[N], b[N];
bool on_diam[N];
int D1, D2, ret;

inline void addedge(int u, int v) {
	to[++E] = v, next[E] = first[u], first[u] = E, ++deg[u];
	to[++E] = u, next[E] = first[v], first[v] = E, ++deg[v];
}

void dfs(int x) {
	int i, y;
	if (dep[x] > dep[ret]) ret = x;
	for (i = first[x]; i; i = next[i])
		if ((y = to[i]) != p[x])
			p[y] = x, dep[y] = dep[x] + 1, dfs(y);
}

int main() {
	int i, j, u, v, s = 0, *ans;
	scanf("%d", &n);
	for (i = 1; i < n; ++i) scanf("%d%d", &u, &v), addedge(u, v);
	ret = 1,  dep[ret] = p[ret] = 0, dfs(ret), D1 = ret;
	ret = D1, dep[ret] = p[ret] = 0, dfs(ret), D2 = ret;
	for (u = D2; u; u = p[u]) on_diam[u] = true;
	for (i = 1; i <= n; ++i) if (!(on_diam[i] || on_diam[p[i]])) return puts("-1"), 0;
	for (u = D2; u; u = p[u]) a[cnt++] = s += std::max(deg[u] - 1, 1);
	assert(s == n);
	for (i = 0; i < cnt - 1; ++i) b[i] = n - a[cnt - 2 - i], b[cnt - 1] = n;
	ans = (std::lexicographical_compare(a, a + cnt, b, b + cnt) ? a : b);
	putchar(49);
	for (i = 0; i < cnt - 1; printf(" %d", ans[i++] + 1))
		for (j = ans[i] + 2; j <= ans[i + 1]; ++j) printf(" %d", j);
	putchar(10);
	return 0;
}