#include<cstdio>
#include<algorithm>
#include<vector>
#define N_ 101000
using namespace std;
int n, v[N_], R1[N_], R2[N_];
vector<int>E[N_], G[N_], U;
int main() {
	int i, a, b, st, j;
	scanf("%d", &n);
	for (i = 1; i < n; i++) {
		scanf("%d%d", &a, &b);
		E[a].push_back(b);
		E[b].push_back(a);
	}
	if (n == 2) {
		printf("%d %d\n", 1, 2);
		return 0;
	}
	for (i = 1; i <= n; i++) {
		if (E[i].size() >= 2) {
			for (auto &t : E[i]) {
				if (E[t].size() >= 2)G[i].push_back(t);
			}
			if (G[i].size() > 2) {
				puts("-1\n");
				return 0;
			}
			if (G[i].size() <= 1) {
				st = i;
			}
		}
	}
	int x = 0;
	for (auto &t : E[st]) {
		if (E[t].size() == 1) {
			x = t;
			break;
		}
	}
	while (1) {
		int nxt = -1;
		int Mx = 0, c = 0;
		v[x] = 1;
		for (auto &t : E[x]) {
			if (!v[t] && E[t].size() > Mx)nxt = t, Mx = E[t].size();
		}
		for (auto &t : E[x]) {
			if (!v[t] && t != nxt)c++;
		}
		U.push_back(c);
		if (nxt == -1)break;
		x = nxt;
	}
	int p1 = 0;
	for (i = 0; i < U.size(); i++) {
		int t = U[i]+1;
		for (j = p1 + 1; j < p1 + t; j++)R1[j] = j+1;
		R1[p1 + t] = p1+1;
		p1 += t;
	}
	p1 = 0;
	for (i = U.size() - 1; i >= 0; i--) {
		int t = U[i]+1;
		for (j = p1 + 1; j < p1 + t; j++)R2[j] = j + 1;
		R2[p1 + t] = p1+1;
		p1 += t;
	}
	int ck = 0;
	for (i = 1; i <= n; i++)if (R1[i] != R2[i]) {
		if (R1[i] > R2[i])for (j = 1; j <= n; j++)R1[j] = R2[j];
		break;
	}
	for (i = 1; i <= n; i++)printf("%d ", R1[i]);
}
### Prompt

Please create a solution in cpp 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<cstdio>
#include<algorithm>
#include<vector>
#define N_ 101000
using namespace std;
int n, v[N_], R1[N_], R2[N_];
vector<int>E[N_], G[N_], U;
int main() {
	int i, a, b, st, j;
	scanf("%d", &n);
	for (i = 1; i < n; i++) {
		scanf("%d%d", &a, &b);
		E[a].push_back(b);
		E[b].push_back(a);
	}
	if (n == 2) {
		printf("%d %d\n", 1, 2);
		return 0;
	}
	for (i = 1; i <= n; i++) {
		if (E[i].size() >= 2) {
			for (auto &t : E[i]) {
				if (E[t].size() >= 2)G[i].push_back(t);
			}
			if (G[i].size() > 2) {
				puts("-1\n");
				return 0;
			}
			if (G[i].size() <= 1) {
				st = i;
			}
		}
	}
	int x = 0;
	for (auto &t : E[st]) {
		if (E[t].size() == 1) {
			x = t;
			break;
		}
	}
	while (1) {
		int nxt = -1;
		int Mx = 0, c = 0;
		v[x] = 1;
		for (auto &t : E[x]) {
			if (!v[t] && E[t].size() > Mx)nxt = t, Mx = E[t].size();
		}
		for (auto &t : E[x]) {
			if (!v[t] && t != nxt)c++;
		}
		U.push_back(c);
		if (nxt == -1)break;
		x = nxt;
	}
	int p1 = 0;
	for (i = 0; i < U.size(); i++) {
		int t = U[i]+1;
		for (j = p1 + 1; j < p1 + t; j++)R1[j] = j+1;
		R1[p1 + t] = p1+1;
		p1 += t;
	}
	p1 = 0;
	for (i = U.size() - 1; i >= 0; i--) {
		int t = U[i]+1;
		for (j = p1 + 1; j < p1 + t; j++)R2[j] = j + 1;
		R2[p1 + t] = p1+1;
		p1 += t;
	}
	int ck = 0;
	for (i = 1; i <= n; i++)if (R1[i] != R2[i]) {
		if (R1[i] > R2[i])for (j = 1; j <= n; j++)R1[j] = R2[j];
		break;
	}
	for (i = 1; i <= n; i++)printf("%d ", R1[i]);
}
```