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<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

int N;
VI G[100011];

VI ord;
int D[100011];
int par[100011];
int bfs(int s) {
    ord.clear();
    ord.push_back(s);
    memset(D, -1, sizeof D);
    D[s] = 0;
    par[s] = s;
    REP (i_, N) {
	int v = ord[i_];
	EACH (e, G[v]) if (D[*e] == -1) {
	    D[*e] = D[v] + 1;
	    par[*e] = v;
	    ord.push_back(*e);
	}
    }

    return max_element(D, D+N) - D;
}

bool online[100011];

void MAIN() {
    scanf("%d", &N);
    REP (i, N-1) {
	int x, y;
	scanf("%d%d", &x, &y);
	x--; y--;
	G[x].push_back(y);
	G[y].push_back(x);
    }

    const int x = bfs(0);
    const int y = bfs(x);

    int cur = y;
    VI vs;
    while (1) {
	online[cur] = true;
	vs.push_back(cur);
	if (cur == x) break;
	cur = par[cur];
    }

    REP (i, N) if (!online[i] && (int)G[i].size() != 1) {
	puts("-1");
	return;
    }

    int cnt = 1;
    VI ans1, ans2;
    EACH (e, vs) {
	int tmp = cnt++;
	REP (t, (int)G[*e].size() - 2) {
	    ans1.push_back(cnt++);
	}
	ans1.push_back(tmp);
    }
    cnt = 1;
    reverse(vs.begin(), vs.end());
    EACH (e, vs) {
	int tmp = cnt++;
	REP (t, (int)G[*e].size() - 2) {
	    ans2.push_back(cnt++);
	}
	ans2.push_back(tmp);
    }

    VI ans = min(ans1, ans2);
    rprintf("%d", ans.begin(), ans.end());
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}

