#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5;
int n;
int sz[maxn + 10], deg[maxn + 10], tot[maxn + 10];
vector<int> g[maxn + 10];
vector<int> a, b, ans;

void dfs(int p, int fa) {
	a.push_back(tot[p]);
	for (auto i: g[p])
		if (sz[i] > 1 && i != fa)
			dfs(i, p);
}

void appendfr() {
	a.push_back(0);
	for (int i = (int)a.size() - 1; i; --i) a[i] = a[i - 1];
	--a[1]; a[0] = 0;
}

void appendbk() {
	--a[(int)a.size() - 1];
	a.push_back(0);
}

void solve() {
	int p = 0, cnt = 0;
	vector<int> now(n);
	for (auto i: a) {
		now[p + i] = ++cnt;
		for (int j = 0; j < i; ++j)
			now[p + j] = ++cnt;
		p += i + 1;
	}
	ans = min(ans, now);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		int l, r; scanf("%d%d", &l, &r);
		g[l].push_back(r); g[r].push_back(l);
		++sz[l]; ++sz[r];
	}
	if (n == 2) {
		printf("1 2\n"); return 0;
	}
	for (int i = 1; i <= n; ++i)
			for (auto j: g[i])
				if (sz[j] > 1) {
					if (sz[i] > 1) ++deg[j];
					else ++tot[j];
				}
	for (int i = 1; i <= n; ++i)
		if (sz[i] > 1 && deg[i] > 2) {
			printf("-1"); return 0;
		}
	for (int i = 1; i <= n; ++i)
		if (sz[i] > 1 && deg[i] < 2) {
			dfs(i, 0); break;
		}
	b = a; ans.push_back(1e9);
	for (int i = 0; i < 2; ++i)
		for (int j = 0; j < 2; ++j) {
			a = b;
			if (i) appendfr();
			if (j) appendbk();
			solve();
			reverse(a.begin(), a.end());
			solve();
		}
	for (auto i: ans) printf("%d ", i);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int maxn = 1e5;
int n;
int sz[maxn + 10], deg[maxn + 10], tot[maxn + 10];
vector<int> g[maxn + 10];
vector<int> a, b, ans;

void dfs(int p, int fa) {
	a.push_back(tot[p]);
	for (auto i: g[p])
		if (sz[i] > 1 && i != fa)
			dfs(i, p);
}

void appendfr() {
	a.push_back(0);
	for (int i = (int)a.size() - 1; i; --i) a[i] = a[i - 1];
	--a[1]; a[0] = 0;
}

void appendbk() {
	--a[(int)a.size() - 1];
	a.push_back(0);
}

void solve() {
	int p = 0, cnt = 0;
	vector<int> now(n);
	for (auto i: a) {
		now[p + i] = ++cnt;
		for (int j = 0; j < i; ++j)
			now[p + j] = ++cnt;
		p += i + 1;
	}
	ans = min(ans, now);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		int l, r; scanf("%d%d", &l, &r);
		g[l].push_back(r); g[r].push_back(l);
		++sz[l]; ++sz[r];
	}
	if (n == 2) {
		printf("1 2\n"); return 0;
	}
	for (int i = 1; i <= n; ++i)
			for (auto j: g[i])
				if (sz[j] > 1) {
					if (sz[i] > 1) ++deg[j];
					else ++tot[j];
				}
	for (int i = 1; i <= n; ++i)
		if (sz[i] > 1 && deg[i] > 2) {
			printf("-1"); return 0;
		}
	for (int i = 1; i <= n; ++i)
		if (sz[i] > 1 && deg[i] < 2) {
			dfs(i, 0); break;
		}
	b = a; ans.push_back(1e9);
	for (int i = 0; i < 2; ++i)
		for (int j = 0; j < 2; ++j) {
			a = b;
			if (i) appendfr();
			if (j) appendbk();
			solve();
			reverse(a.begin(), a.end());
			solve();
		}
	for (auto i: ans) printf("%d ", i);
}

```