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;

const int maxn = 1e5 + 5;

int n;
vector<int> g[maxn];
int d[maxn], vis[maxn];

void dfs(int u, int par) {
	d[u] = par ? d[par] + 1 : 0;
	vis[u] = 1;
	for(int v : g[u]) if(!vis[v]) {
		dfs(v, u);
	}
}
int s, t;
vector<int> dia;
int get_dia(int u, int par) {
	if(u == t) {
		dia.emplace_back(u);
		return 1;
	}
	for(int v : g[u]) if(v - par) {
		if(get_dia(v, u)) {
			dia.emplace_back(u);
			return 1;
		}
	} return 0;
}

vector<int> solve() {
	vector<int> k;
	k.emplace_back(0);
	for(int i = 1; i < (int)dia.size() - 1; i++) {
		int u = dia[i];
		int cnt = 0;
		for(int v : g[u]) {
			if(v != dia[i - 1] and v != dia[i + 1]) {
				cnt++;
			}
		}
		k.emplace_back(cnt);
	}
	k.emplace_back(0);
	int c = 0;
	vector<int> ret(n);
	iota(ret.begin(), ret.end(), 1);
	for(int i : k) {
		if(i) {
			rotate(ret.begin() + c, ret.begin() + c + i, ret.begin() + c + i + 1);
		}
		c += i + 1;
	}
	return ret;
}

int main(int argc, char const *argv[])
{
	// freopen("in", "r", stdin);
	cin >> n;
	for(int i = 1; i < n; i++) {
		int u, v;
		cin >> u >> v;
		g[u].emplace_back(v);
		g[v].emplace_back(u);
	}
	memset(vis, 0, sizeof vis);
	dfs(1, 0);
	s = 0, t = 0;
	for(int i = 1; i <= n; i++) {
		if(d[i] > d[s]) {
			s = i;
		}
	}
	memset(vis, 0, sizeof vis);
	dfs(s, 0);
	for(int i = 1; i <= n; i++) {
		if(d[i] > d[t]) {
			t = i;
		}
	}
	get_dia(s, 0);

	memset(vis, 0, sizeof vis);
	queue<int> Q;
	for(int u : dia) {
		vis[u] = 1;
		d[u] = 0;
		Q.emplace(u);
	}
	while(Q.size()) {
		int u = Q.front();
		Q.pop();
		for(int v : g[u]) if(!vis[v]) {
			vis[v] = 1;
			d[v] = d[u] + 1;
		}
	}
	for(int i = 1; i <= n; i++) {
		if(d[i] > 1) {
			cout << "-1" << endl;
			exit(0);
		}
	}
	vector<int> mn(n);
	iota(mn.begin(), mn.end(), 1);
	reverse(mn.begin(), mn.end());
	mn = min(mn, solve());
	reverse(dia.begin(), dia.end());
	mn = min(mn, solve());
	vector<int> p(n);
	for(int i = 0; i < n; i++) {
		p[mn[i] - 1] = i + 1;
	}
	for(int i : p) cout << i << ' '; cout << endl;
	return 0;
}