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>
#include <cstdio>
#include <vector>
using namespace std;

const int N = 100000 + 5;

int n, s, t, d;
vector<int> g[N];

void dfs(int x, int fa, int dis, int &point) {
    if (dis > d) {
        d = dis;
        point = x;
    }
    for (int i = 0; i < g[x].size(); ++i) {
        int y = g[x][i];
        if (y == fa) continue;
        dfs(y, x, dis + 1, point);
    }
}

vector<int> path;
bool color[N];

bool find_path(int x, int fa) {
    path.push_back(x); color[x] = true;
    if (x == t) return true;
    for (int i = 0; i < g[x].size(); ++i) {
        int y = g[x][i];
        if (y == fa) continue;
        if (find_path(y, x)) return true;
    }
    path.pop_back(), color[x] = false;
    return false;
}

bool vis[N];
int cnt[N];
int ans1[N], ans2[N];

int main() {
    scanf("%d", &n);
    for (int i = 1, a, b; i < n; ++i) {
        scanf("%d%d", &a, &b);
        g[a].push_back(b), g[b].push_back(a);
    }
    dfs(1, 0, 0, s), d = 0, dfs(s, 0, 0, t);
    find_path(s, 0);
    for (int i = 0; i < path.size(); ++i) {
        int x = path[i];
        for (int j = 0; j < g[x].size(); ++j) {
            int y = g[x][j];
            vis[y] = true;
            if (!color[y]) ++cnt[i];
        }
    }
    for (int i = 1; i <= n; ++i) if (!vis[i]) { puts("-1"); return 0; }
    int pos = 1, id = 1;
    for (int i = 0; i < path.size(); ++i) {
        for (int j = 1; j <= cnt[i]; ++j)
            ans1[pos++] = id + j;
        ans1[pos++] = id;
        id += cnt[i] + 1;
    }
    pos = 1, id = 1;
    for (int i = path.size() - 1; i >= 0; --i) {
        for (int j = 1; j <= cnt[i]; ++j)
            ans2[pos++] = id + j;
        ans2[pos++] = id;
        id += cnt[i] + 1;
    }
    int t = 1;
    for (int i = 1; i <= n; ++i) if (ans1[i] < ans2[i]) { t = 1; break; } else if (ans1[i] > ans2[i]) { t = 2; break; }
    if (t == 1) for (int i = 1; i <= n; ++i) printf("%d%c", ans1[i], i == n ? '\n' : ' ');
    else for (int i = 1; i <= n; ++i) printf("%d%c", ans2[i], i == n ? '\n' : ' ');
    return 0;
}
