#include <bits/stdc++.h>
using namespace std;

const int N = 100010;

struct edge {
    int to, next;
} e[N * 2];

int head[N], ecnt, ind[N];

inline void adde(int from, int to) {
    e[++ecnt] = (edge){to, head[from]}, head[from] = ecnt;
    e[++ecnt] = (edge){from, head[to]}, head[to] = ecnt;
    ind[from]++, ind[to]++;
}

int dep[N], mx, pos, tmp[N]; vector<int> hh, ans1, ans2;

void dfs(int u, int f) {
    dep[u] = dep[f] + 1, tmp[u] = 0;
    if (dep[u] > mx) mx = dep[u], pos = u;
    for (int i = head[u]; i; i = e[i].next)
        if (e[i].to != f) dfs(e[i].to, u), tmp[u] = max(tmp[u], tmp[e[i].to] + 1);
}

void dfs2(int u, int f) {
    for (int i = head[u]; i; i = e[i].next)
        if (e[i].to != f && tmp[e[i].to] + 1== tmp[u]) {
            dfs2(e[i].to, u); break;
        }
    hh.push_back(ind[u] - (!!f) - (!!tmp[u]) + 1);
}

int main() {
    int n; scanf("%d", &n);
    for (int i = 1, a, b; i < n; i++) 
        scanf("%d%d", &a, &b), adde(a, b);
    dfs(1, 0); int tp = 0, tot = 0, u = pos; ans1.resize(n + 1);
    dfs(u, 0),dfs2(u, 0);
    for (auto t : hh) {
        ans1[tp + t] = ++tot;
        for (int i = tp + 1; i < tp + t; i++) ans1[i] = ++tot;
        tp += t;
    }
    reverse(hh.begin(), hh.end()), ans2.resize(n + 1), tot = 0, tp = 0;
    for (auto t : hh) {
        ans2[tp + t] = ++tot;
        for (int i = tp + 1; i < tp + t; i++) ans2[i] = ++tot;
        tp += t;
    }
    if (ans1 > ans2) swap(ans1, ans2);
    for (int i = 1; i <= n; i++) if (!ans1[i]) 
        return puts("-1"), 0;
    for (int i = 1; i <= n; i++) printf("%d ", ans1[i]);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 <bits/stdc++.h>
using namespace std;

const int N = 100010;

struct edge {
    int to, next;
} e[N * 2];

int head[N], ecnt, ind[N];

inline void adde(int from, int to) {
    e[++ecnt] = (edge){to, head[from]}, head[from] = ecnt;
    e[++ecnt] = (edge){from, head[to]}, head[to] = ecnt;
    ind[from]++, ind[to]++;
}

int dep[N], mx, pos, tmp[N]; vector<int> hh, ans1, ans2;

void dfs(int u, int f) {
    dep[u] = dep[f] + 1, tmp[u] = 0;
    if (dep[u] > mx) mx = dep[u], pos = u;
    for (int i = head[u]; i; i = e[i].next)
        if (e[i].to != f) dfs(e[i].to, u), tmp[u] = max(tmp[u], tmp[e[i].to] + 1);
}

void dfs2(int u, int f) {
    for (int i = head[u]; i; i = e[i].next)
        if (e[i].to != f && tmp[e[i].to] + 1== tmp[u]) {
            dfs2(e[i].to, u); break;
        }
    hh.push_back(ind[u] - (!!f) - (!!tmp[u]) + 1);
}

int main() {
    int n; scanf("%d", &n);
    for (int i = 1, a, b; i < n; i++) 
        scanf("%d%d", &a, &b), adde(a, b);
    dfs(1, 0); int tp = 0, tot = 0, u = pos; ans1.resize(n + 1);
    dfs(u, 0),dfs2(u, 0);
    for (auto t : hh) {
        ans1[tp + t] = ++tot;
        for (int i = tp + 1; i < tp + t; i++) ans1[i] = ++tot;
        tp += t;
    }
    reverse(hh.begin(), hh.end()), ans2.resize(n + 1), tot = 0, tp = 0;
    for (auto t : hh) {
        ans2[tp + t] = ++tot;
        for (int i = tp + 1; i < tp + t; i++) ans2[i] = ++tot;
        tp += t;
    }
    if (ans1 > ans2) swap(ans1, ans2);
    for (int i = 1; i <= n; i++) if (!ans1[i]) 
        return puts("-1"), 0;
    for (int i = 1; i <= n; i++) printf("%d ", ans1[i]);
}

```