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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#define MAXN 100010
int head[MAXN], to[MAXN << 1], next[MAXN << 1], tot = 0;
inline void $(int u, int v) {
    next[tot] = head[u], to[tot] = v, head[u] = tot++;
    next[tot] = head[v], to[tot] = u, head[v] = tot++;
}
int n, dis[MAXN], fa[MAXN];
std::queue<int> Q;
int bfs(int x) {
    Q.push(x);
    dis[x] = fa[x] = 0;
    int ans = x;
    while (Q.size()) {
        int x = Q.front();
        Q.pop();
        for (int i = head[x]; ~i; i = next[i]) {
            if (to[i] == fa[x]) continue;
            fa[to[i]] = x;
            if ((dis[to[i]] = dis[x] + 1) > dis[ans]) ans = to[i];
            Q.push(to[i]);
        }
    }
    return ans;
}
int h, t, body[MAXN], k[MAXN], p[MAXN], m;
int ans0[MAXN], ans1[MAXN];
int main() {
    memset(head, -1, sizeof(head));
    scanf("%d", &n);
    for (int i = 1, x, y; i < n; i++) {
        scanf("%d%d", &x, &y);
        $(x, y);
    }
    t = bfs(h = bfs(1));
    // printf("H : %d, T : %d\n", h, t);
    for (int x = t;; x = fa[x]) {
        body[x] = 1;
        p[++m] = x;
        // printf("%d ", x);
        if (x == h) break;
    }
    // puts("");
    for (int i = 1; i <= n; i++) {
        if (!body[i]) {
            if (!body[fa[i]]) return puts("-1"), 0;
            k[fa[i]]++;
        }
    }
    for (int i = 1, x = 1; i <= m; i++) {
        for (int j = 1; j <= k[p[i]]; j++) ans0[x + j - 1] = x + j;
        ans0[x + k[p[i]]] = x;
        x += k[p[i]] + 1;
    }
    std::reverse(p + 1, p + m + 1);
    for (int i = 1, x = 1; i <= m; i++) {
        for (int j = 1; j <= k[p[i]]; j++) ans1[x + j - 1] = x + j;
        ans1[x + k[p[i]]] = x;
        x += k[p[i]] + 1;
    }
    for (int i = 1; i <= n; i++) {
        if (ans0[i] > ans1[i]) {
            std::swap(ans0, ans1);
            break;
        } else if (ans0[i] < ans1[i])
            break;
    }
    for (int i = 1; i <= n; i++) printf("%d%c", ans0[i], " \n"[i == n]);
    return 0;
}