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<vector>
#include<algorithm>
using namespace std;

const int MX = 100005;
int n, v, u, c = 2;
vector<int> G[MX], a;

int dfs1(int u, int p) {
	int c = u != v, ret = u;
	for (int nx : G[u]) if (nx != p && G[nx].size() > 1) {
		if (c++ > 1) {
			puts("-1");
			exit(0);
		}
		ret = dfs1(nx, u);
	}
	return ret;
}
void dfs2(int u, int p) {
	a.push_back((int)G[u].size());
	for (int nx : G[u]) if (nx != p && G[nx].size() > 1) dfs2(nx, u);
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n - 1; i++) {
		scanf("%d%d", &v, &u);
		v--; u--;
		G[v].push_back(u);
		G[u].push_back(v);
	}
	if (G[v].size() == 1) v = G[v][0];
	v = dfs1(v, -1);
	dfs2(v, -1);
	u = (int)a.size();
	for (int i = 0; i < u; i++) if (a[i] != a[u - 1 - i]) {
		if (a[i] > a[u - 1 - i]) reverse(a.begin(), a.end());
		break;
	}
	printf("1 ");
	for (int i = 0; i < u; i++) {
		int t = c++;
		for (int j = 0; j < a[i] - 2; j++) printf("%d ", c++);
		printf("%d ", t);
		if (i == u - 1) printf("%d\n", c++);
	}

	return 0;
}