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 MN = 100010;

int N;
vector<int> adj[MN];
int dist[MN], par[MN];

void dfs0(int u, int p) {
    for(int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i];
        if(v == p) continue;
        dist[v] = dist[u] + 1;
        par[v] = u;
        dfs0(v, u);
    }
}

vector<int> V;
int del[MN], sz[MN];

void dfs1(int u, int p) {
    sz[u] = 1;
    for(int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i];
        if(v == p) continue;
        if(del[v]) continue;
        dist[v] = dist[u] + 1;
        dfs1(v, u);
        sz[u] += sz[v];
    }
}

bool cmp(vector<int> &a, vector<int> &b) {
    for(int i = 0; i < a.size(); i++) {
        if(a[i] != b[i]) return a[i] < b[i];
    }
    return false;
}

int main() {
    scanf("%d", &N);

    for(int i = 0; i < N - 1; i++) {
        int u, v; scanf("%d %d", &u, &v);
        u--; v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs0(0, -1);

    int p = 0;
    for(int i = 0; i < N; i++) if(dist[i] > dist[p]) p = i;

    dist[p] = 0;
    par[p] = -1;
    dfs0(p, -1);

    int q = 0;
    for(int i = 0; i < N; i++) if(dist[i] > dist[q]) q = i;

    int u = q;
    while(u != -1) {
        V.push_back(u);
        del[u] = 1;
        u = par[u];
    }

    for(int i = 0; i < V.size(); i++) {
        dist[ V[i] ] = 0;
        dfs1(V[i], -1);
    }

    for(int i = 0; i < N; i++) if(dist[i] > 1) {
        printf("-1");
        return 0;
    }

    vector<int> A, B;
    for(int i = 0; i < V.size(); i++) A.push_back(sz[ V[i] ] - 1);
    for(int i = (int)V.size() - 1; i >= 0; i--) B.push_back(sz[ V[i] ] - 1);
    if(!cmp(A, B)) swap(A, B);

    vector<int> ans;
    int pos = 1;
    for(int i = 0; i < A.size(); i++) {
        for(int j = pos + 1; j <= pos + A[i]; j++) {
            ans.push_back(j);
        }
        ans.push_back(pos);
        pos += A[i] + 1;
    }

    for(int i = 0; i < N; i++) {
        printf("%d ", ans[i]);
    }
}
