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 <cstring>

using namespace std;

const int N = 100000;

int n, num, head[N + 50], deep[N + 50], rot1, rot2, f[N + 50], on[N + 50], ans1[N + 50], ans2[N + 50], cnt, nxt[N + 50], ctt, son[N + 50];

struct Node
{
    int next, to;
} edge[N * 2 + 50];

void Addedge(int u, int v)
{
    edge[++num] = (Node){head[u], v};
    head[u] = num;
    return;
}

int Dfs(int x, int fa)
{
    int maxx = 0, bh = x;
    deep[x] = deep[fa] + 1; f[x] = fa;
    for (int i = head[x]; i; i = edge[i].next)
    {
        int v = edge[i].to;
        if (v == fa) continue;
        int tmp = Dfs(v, x); 
        if (deep[tmp] > maxx) maxx = deep[tmp], bh = tmp, nxt[x] = v;
    }
    return bh;
}

int main()
{
    scanf("%d", &n);
    for (int i = 1, u, v; i <= n - 1; i++)
    {
        scanf("%d%d", &u, &v);
        Addedge(u, v); Addedge(v, u);
    }
    rot1 = Dfs(1, 0);
    for (int i = 1; i <= n; i++) deep[i] = 0, f[i] = 0, nxt[i] = 0;
    rot2 = Dfs(rot1, 0);
    int tmp = rot1;
    while (tmp) on[tmp] = 1, tmp = nxt[tmp];
    for (int i = 1; i <= n; i++)
    {
        if (on[i]) continue;
        int flag = 0;
        for (int j = head[i]; j; j = edge[j].next)	
        {
            int v = edge[j].to;
            if (on[v]) { son[v]++; flag = 1; break; } 
        }
        if (!flag) { puts("-1"); return 0; }
    }
    for(int i = rot2; i; i = f[i])
    {
        ans1[ctt + son[i] + 1] = ++cnt;
        for (int j = 1; j <= son[i]; j++) ans1[++ctt] = ++cnt;
        ctt++;
    }
    ctt = cnt = 0;
    for (int i = rot1; i; i = nxt[i])
    {
    	ans2[ctt + son[i] + 1] = ++cnt;
    	for (int j = 1; j <= son[i]; j++) ans2[++ctt] = ++cnt;
    	ctt++;
	}
	for (int i = 1; i <= ctt; i++)
		if (ans1[i] != ans2[i])
		{
			if (ans1[i] < ans2[i]) for (int i = 1; i <= n; i++) printf("%d ", ans1[i]);
			else for (int i = 1; i <= n; i++) printf("%d ", ans2[i]);
			return 0;
		}
	for (int i = 1; i <= n; i++) printf("%d ", ans1[i]);
    return 0; 
}