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>
using namespace std;
int n, c, a[100009], b[100009], deg[100009], deg2[100009], g[100009][2], v[100009], ans[100009], cur[100009];
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		cin >> a[i] >> b[i];
		deg[--a[i]]++;
		deg[--b[i]]++;
	}
	bool ok = true;
	for (int i = 0; i < n - 1; i++) {
		if (deg[a[i]] >= 2 && deg[b[i]] >= 2) {
			if (deg2[a[i]] <= 1) g[a[i]][deg2[a[i]]++] = b[i];
			else ok = false;
			if (deg2[b[i]] <= 1) g[b[i]][deg2[b[i]]++] = a[i];
			else ok = false;
		}
	}
	int m = 0;
	for (int i = 0; i < n; i++) {
		if (deg2[i] >= 1) m++;
	}
	if (m == 0) {
		if (n >= 3) {
			cout << '1';
			for (int i = 3; i <= n - 1; i++) cout << ' ' << i;
			cout << " 2 " << n << "\n";
		}
		else cout << "1 2\n";
	}
	else if (!ok) cout << "-1\n";
	else {
		ans[0] = 1 << 30;
		for (int i = 0; i < n; i++) {
			if (deg2[i] != 1) continue;
			int pos = i, pre = -1, cnt = 0;
			for (int j = 0; j < m; j++) {
				cur[cnt++] = deg[pos] - deg2[pos] + (j != 0 && j != m - 1);
				for (int k = 0; k < deg2[pos]; k++) {
					if (g[pos][k] != pre) {
						pre = pos;
						pos = g[pos][k];
						break;
					}
				}
			}
			for (int j = 0; j < m; j++) {
				if (ans[j] < cur[j]) break;
				else if (ans[j] > cur[j]) {
					for (int k = 0; k < m; k++) {
						ans[k] = cur[k];
					}
					break;
				}
			}
		}
		cout << 1;
		int s = 1;
		for (int i = 0; i < m; i++) {
			for (int j = s + 2; j <= s + ans[i]; j++) cout << ' ' << j;
			cout << ' ' << s + 1;
			s += ans[i];
		}
		cout << ' ' << n << '\n';
	}
	return 0;
}