#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 100010;

int n;
int sz[MN];
vi g[MN], g2[MN];
bool f[MN], vis[MN];

void ng()
{
	puts("-1");
	exit(0);
}

void show(vi vec)
{
	rep(i, vec.size()) {
		printf("%d%c", vec[i], i+1 == vec.size() ? '\n' : ' ');
	}
}

vi solve(vi deg)
{
	vi res;
	res.pb(1);
	int now = 1;

	rep(i, deg.size()) {
		int x = deg[i];
		if (i == 0) --x;
		if (i+1 == (int)deg.size()) --x;

		int nx = ++now;

		while (x) {
			res.pb(++now);
			--x;
		}
		res.pb(nx);
	}

	res.pb(++now);
	//show(res);
	return res;
}

int main() {
	cin >> n;

	rep(i, n-1) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		g[a].pb(b); g[b].pb(a);
		++sz[a]; ++sz[b];
	}

	if (n == 2) {
		puts("1 2");
		return 0;
	}

	rep(i, n) {
		for (int to : g[i]) {
			if (sz[i] >= 2 && sz[to] >= 2 && i < to) {
				g2[i].pb(to);
				g2[to].pb(i);
			}
		}
	}

	int cnt = 0;

	rep(i, n) {
		if (g2[i].size() > 2) {
			ng();
		}
		if (g2[i].size() > 0) {
			++cnt;
		}
	}
	//printf("cnt=%d\n", cnt);

	vi ans;

	rep(i, n) if (g2[i].size() == 1) {
		vi vs;
		int c = i;

		while (true) {
			bool fd = 0;
			vs.pb(c);
			vis[c] = 1;

			for (int to : g2[c]) {
				if (!vis[to]) {
					fd = 1;
					c = to;
					break;
				}
			}
			if (!fd) {
				break;
			}
		}

		if ((int)vs.size() != cnt) {
			ng();
		}

		vi vec;

		rep(i, vs.size()) {
			vec.pb((int)g[vs[i]].size() - (int)g2[vs[i]].size());
		}

		vi cand = solve(vec);
		reverse(ALL(vec));

		vi cand2 = solve(vec);
		show(min(cand, cand2));
		return 0;
	}

	rep(i, n) {
		if (sz[i] >= 2) {
			vi vec;
			vec.pb(sz[i]);
			vi t = solve(vec);
			show(t);
			return 0;
		}
	}

	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 100010;

int n;
int sz[MN];
vi g[MN], g2[MN];
bool f[MN], vis[MN];

void ng()
{
	puts("-1");
	exit(0);
}

void show(vi vec)
{
	rep(i, vec.size()) {
		printf("%d%c", vec[i], i+1 == vec.size() ? '\n' : ' ');
	}
}

vi solve(vi deg)
{
	vi res;
	res.pb(1);
	int now = 1;

	rep(i, deg.size()) {
		int x = deg[i];
		if (i == 0) --x;
		if (i+1 == (int)deg.size()) --x;

		int nx = ++now;

		while (x) {
			res.pb(++now);
			--x;
		}
		res.pb(nx);
	}

	res.pb(++now);
	//show(res);
	return res;
}

int main() {
	cin >> n;

	rep(i, n-1) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		g[a].pb(b); g[b].pb(a);
		++sz[a]; ++sz[b];
	}

	if (n == 2) {
		puts("1 2");
		return 0;
	}

	rep(i, n) {
		for (int to : g[i]) {
			if (sz[i] >= 2 && sz[to] >= 2 && i < to) {
				g2[i].pb(to);
				g2[to].pb(i);
			}
		}
	}

	int cnt = 0;

	rep(i, n) {
		if (g2[i].size() > 2) {
			ng();
		}
		if (g2[i].size() > 0) {
			++cnt;
		}
	}
	//printf("cnt=%d\n", cnt);

	vi ans;

	rep(i, n) if (g2[i].size() == 1) {
		vi vs;
		int c = i;

		while (true) {
			bool fd = 0;
			vs.pb(c);
			vis[c] = 1;

			for (int to : g2[c]) {
				if (!vis[to]) {
					fd = 1;
					c = to;
					break;
				}
			}
			if (!fd) {
				break;
			}
		}

		if ((int)vs.size() != cnt) {
			ng();
		}

		vi vec;

		rep(i, vs.size()) {
			vec.pb((int)g[vs[i]].size() - (int)g2[vs[i]].size());
		}

		vi cand = solve(vec);
		reverse(ALL(vec));

		vi cand2 = solve(vec);
		show(min(cand, cand2));
		return 0;
	}

	rep(i, n) {
		if (sz[i] >= 2) {
			vi vec;
			vec.pb(sz[i]);
			vi t = solve(vec);
			show(t);
			return 0;
		}
	}

	return 0;
}
```