#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long

const int maxN = 100011;

int n, i, j, x, y, bg, pos;
vector<int> list[maxN];
bool leaf[maxN];
int gr[maxN], gr_leaf[maxN];
vector<int> ord, ord2;
vector<int> ans;

void dfs(int node, int root) {
    ord.pb(node);
    for (auto to : list[node])
        if (!leaf[to])
            if (to != root)
                dfs(to, node);
}

int main()
{
   // freopen("test.in","r",stdin);

    scanf("%d", &n);
    for (i = 1; i < n; i++) {
        scanf("%d%d", &x, &y);
        list[x].pb(y);
        list[y].pb(x);
        gr[x]++; gr[y]++;
    }

    if (n == 2) {
        printf("1 2");
        return 0;
    }

    for (i = 1; i <= n; i++) {
        if (gr[i] > 1) continue;
        leaf[i] = true;
        gr_leaf[list[i][0]]++;
    }

    for (i = 1; i <= n; i++) {
        if (gr[i] - gr_leaf[i] > 2) {
            printf("-1");
            return 0;
        }
    }

    for (bg = 1; leaf[bg] || gr[bg] - gr_leaf[bg] == 2; bg++);
    dfs(bg, 0);

    ord2 = ord;
    reverse(ord2.begin(), ord2.end());
    for (i = 0; i < ord.size(); i++) {

        if (gr_leaf[ord[i]] < gr_leaf[ord2[i]]) break;
        if (gr_leaf[ord[i]] > gr_leaf[ord2[i]]) {
            swap(ord, ord2);
            break;
        }
    }

    gr_leaf[ord[0]]--;
    gr_leaf[ord.back()]--;

    pos = 2;
    ans = {1};

    for (auto e : ord) {
        for (i = 1; i <= gr_leaf[e]; i++) ans.pb(pos + i);
        ans.pb(pos);
        pos += gr_leaf[e] + 1;
    }

    ans.pb(n);

    for (auto e : ans) printf("%d ", e);


    return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long

const int maxN = 100011;

int n, i, j, x, y, bg, pos;
vector<int> list[maxN];
bool leaf[maxN];
int gr[maxN], gr_leaf[maxN];
vector<int> ord, ord2;
vector<int> ans;

void dfs(int node, int root) {
    ord.pb(node);
    for (auto to : list[node])
        if (!leaf[to])
            if (to != root)
                dfs(to, node);
}

int main()
{
   // freopen("test.in","r",stdin);

    scanf("%d", &n);
    for (i = 1; i < n; i++) {
        scanf("%d%d", &x, &y);
        list[x].pb(y);
        list[y].pb(x);
        gr[x]++; gr[y]++;
    }

    if (n == 2) {
        printf("1 2");
        return 0;
    }

    for (i = 1; i <= n; i++) {
        if (gr[i] > 1) continue;
        leaf[i] = true;
        gr_leaf[list[i][0]]++;
    }

    for (i = 1; i <= n; i++) {
        if (gr[i] - gr_leaf[i] > 2) {
            printf("-1");
            return 0;
        }
    }

    for (bg = 1; leaf[bg] || gr[bg] - gr_leaf[bg] == 2; bg++);
    dfs(bg, 0);

    ord2 = ord;
    reverse(ord2.begin(), ord2.end());
    for (i = 0; i < ord.size(); i++) {

        if (gr_leaf[ord[i]] < gr_leaf[ord2[i]]) break;
        if (gr_leaf[ord[i]] > gr_leaf[ord2[i]]) {
            swap(ord, ord2);
            break;
        }
    }

    gr_leaf[ord[0]]--;
    gr_leaf[ord.back()]--;

    pos = 2;
    ans = {1};

    for (auto e : ord) {
        for (i = 1; i <= gr_leaf[e]; i++) ans.pb(pos + i);
        ans.pb(pos);
        pos += gr_leaf[e] + 1;
    }

    ans.pb(n);

    for (auto e : ans) printf("%d ", e);


    return 0;
}

```