#include<bits/stdc++.h>
using namespace std;
int n, fa[100007], dep[100007], d[100007], tot, cnt[100007], res1[100007], res2[100007];
vector<int>e[100007];
bool is[100007];
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int w, int f)
{
	fa[w] = f;
	dep[w] = dep[f] + 1;
	int res = w;
	for (int i = 0, lim = e[w].size(); i < lim; i++)
		if (e[w][i] != f)
		{
			int ret = dfs(e[w][i], w);
			if (dep[ret] > dep[res])res = ret;
		}
	return res;
}
int print(int* p)
{
	for (int i = 1; i <= n; i++)
		printf("%d ", p[i]);
	putchar('\n');
	return 0;
}
int main()
{
	n = read();
	for (int i = 2; i <= n; i++)
	{
		int a = read(), b = read();
		e[a].push_back(b);
		e[b].push_back(a);
	}
	int mx = dfs(1, 0);
	mx = dfs(mx, 0);
	while (mx)d[++tot] = mx, is[mx] = 1, mx = fa[mx];
	for (int i = 1; i <= n; i++)
		if (!is[i])
		{
			bool ok = 0;
			for (int j = 0, lim = e[i].size(); j < lim && !ok; j++)
				if (is[e[i][j]])cnt[e[i][j]]++, ok = 1;
			if (!ok) { cout << -1 << endl; return 0; }
		}
	int pos = 0;
	for (int i = 1; i <= tot; i++)
	{
		for (int j = 1; j <= cnt[d[i]]; j++)
			res1[pos + j] = pos + j + 1;
		res1[pos + cnt[d[i]] + 1] = pos + 1;
		pos += cnt[d[i]] + 1;
	}
	pos = 0;
	for (int i = tot; i >= 1; i--)
	{
		for (int j = 1; j <= cnt[d[i]]; j++)
			res2[pos + j] = pos + j + 1;
		res2[pos + cnt[d[i]] + 1] = pos + 1;
		pos += cnt[d[i]] + 1;
	}
	for (int i = 1; i <= n; i++)
		if (res1[i] > res2[i]) { print(res2); return 0; }
		else if (res1[i] < res2[i]) { print(res1); return 0; }
	print(res1);
	return 0;
}
### Prompt

Create a solution in cpp for 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<bits/stdc++.h>
using namespace std;
int n, fa[100007], dep[100007], d[100007], tot, cnt[100007], res1[100007], res2[100007];
vector<int>e[100007];
bool is[100007];
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int w, int f)
{
	fa[w] = f;
	dep[w] = dep[f] + 1;
	int res = w;
	for (int i = 0, lim = e[w].size(); i < lim; i++)
		if (e[w][i] != f)
		{
			int ret = dfs(e[w][i], w);
			if (dep[ret] > dep[res])res = ret;
		}
	return res;
}
int print(int* p)
{
	for (int i = 1; i <= n; i++)
		printf("%d ", p[i]);
	putchar('\n');
	return 0;
}
int main()
{
	n = read();
	for (int i = 2; i <= n; i++)
	{
		int a = read(), b = read();
		e[a].push_back(b);
		e[b].push_back(a);
	}
	int mx = dfs(1, 0);
	mx = dfs(mx, 0);
	while (mx)d[++tot] = mx, is[mx] = 1, mx = fa[mx];
	for (int i = 1; i <= n; i++)
		if (!is[i])
		{
			bool ok = 0;
			for (int j = 0, lim = e[i].size(); j < lim && !ok; j++)
				if (is[e[i][j]])cnt[e[i][j]]++, ok = 1;
			if (!ok) { cout << -1 << endl; return 0; }
		}
	int pos = 0;
	for (int i = 1; i <= tot; i++)
	{
		for (int j = 1; j <= cnt[d[i]]; j++)
			res1[pos + j] = pos + j + 1;
		res1[pos + cnt[d[i]] + 1] = pos + 1;
		pos += cnt[d[i]] + 1;
	}
	pos = 0;
	for (int i = tot; i >= 1; i--)
	{
		for (int j = 1; j <= cnt[d[i]]; j++)
			res2[pos + j] = pos + j + 1;
		res2[pos + cnt[d[i]] + 1] = pos + 1;
		pos += cnt[d[i]] + 1;
	}
	for (int i = 1; i <= n; i++)
		if (res1[i] > res2[i]) { print(res2); return 0; }
		else if (res1[i] < res2[i]) { print(res1); return 0; }
	print(res1);
	return 0;
}
```