#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define FOR(i, a, b) for(int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for(int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define all(v) v.begin(), v.end()
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

int n, l, a, b, p[100005], ans[2][100005], cur;
pi h, k;
bool diam[100005];
vector<int> adj[100005], v;

void dfs(int x, int par, int d) {
	k = max(k, pi(d, x));
	p[x] = par;
	for (auto it:adj[x]) {
		if (it == par) continue;
		dfs(it, x, d+1);
	}
}

int main() {
	fastio; cin >> n;
	FOR(i, 1, n-1) {
		cin >> a >> b;
		adj[a].pb(b);
		adj[b].pb(a);
	}
	dfs(1, -1, 0); swap(h, k);
	dfs(h.s, -1, 0);
	a = h.s, b = k.s; //endpoints of diam are a, b
	
	v.pb(adj[b].size()-1);
	diam[b] = 1;
	while (b != a) {
		b = p[b]; diam[b] = 1;
		v.pb(adj[b].size()-2);
		if (b == a) v.back()++;
	}
	FOR(i, 1, n) if (!diam[i]) {
		bool ok = false;
		for (auto it:adj[i]) if (diam[it]) ok = 1;
		if (!ok) {cout << -1; return 0;}
	}
	
	l = v.size();
	FOR(i, 1, n) ans[0][i] = ans[1][i] = i;
	FOR(i, 0, l-1) {
		if (v[i] == 0) continue;
		FOR(j, 0, v[i]-1) swap(ans[0][cur+i+1+j], ans[0][cur+i+1+j+1]);
		//shift cur + i + 1 to v[i] positions to the right
		cur += v[i];
	}
	reverse(all(v)); cur = 0;
	FOR(i, 0, l-1) {
		if (v[i] == 0) continue;
		FOR(j, 0, v[i]-1) swap(ans[1][cur+i+1+j], ans[1][cur+i+1+j+1]);
		//shift cur + i + 1 to v[i] positions to the right
		cur += v[i];
	}
	FOR(i, 1, n) if (ans[0][i] != ans[1][i]) {
		FOR(j, 1, n) cout << ans[(ans[0][i] < ans[1][i] ? 0 : 1)][j] << " ";
		return 0;
	}
	FOR(i, 1, n) cout << ans[0][i] << " ";
}

### 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 long long ll;
typedef long double ld;
#define FOR(i, a, b) for(int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for(int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define all(v) v.begin(), v.end()
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

int n, l, a, b, p[100005], ans[2][100005], cur;
pi h, k;
bool diam[100005];
vector<int> adj[100005], v;

void dfs(int x, int par, int d) {
	k = max(k, pi(d, x));
	p[x] = par;
	for (auto it:adj[x]) {
		if (it == par) continue;
		dfs(it, x, d+1);
	}
}

int main() {
	fastio; cin >> n;
	FOR(i, 1, n-1) {
		cin >> a >> b;
		adj[a].pb(b);
		adj[b].pb(a);
	}
	dfs(1, -1, 0); swap(h, k);
	dfs(h.s, -1, 0);
	a = h.s, b = k.s; //endpoints of diam are a, b
	
	v.pb(adj[b].size()-1);
	diam[b] = 1;
	while (b != a) {
		b = p[b]; diam[b] = 1;
		v.pb(adj[b].size()-2);
		if (b == a) v.back()++;
	}
	FOR(i, 1, n) if (!diam[i]) {
		bool ok = false;
		for (auto it:adj[i]) if (diam[it]) ok = 1;
		if (!ok) {cout << -1; return 0;}
	}
	
	l = v.size();
	FOR(i, 1, n) ans[0][i] = ans[1][i] = i;
	FOR(i, 0, l-1) {
		if (v[i] == 0) continue;
		FOR(j, 0, v[i]-1) swap(ans[0][cur+i+1+j], ans[0][cur+i+1+j+1]);
		//shift cur + i + 1 to v[i] positions to the right
		cur += v[i];
	}
	reverse(all(v)); cur = 0;
	FOR(i, 0, l-1) {
		if (v[i] == 0) continue;
		FOR(j, 0, v[i]-1) swap(ans[1][cur+i+1+j], ans[1][cur+i+1+j+1]);
		//shift cur + i + 1 to v[i] positions to the right
		cur += v[i];
	}
	FOR(i, 1, n) if (ans[0][i] != ans[1][i]) {
		FOR(j, 1, n) cout << ans[(ans[0][i] < ans[1][i] ? 0 : 1)][j] << " ";
		return 0;
	}
	FOR(i, 1, n) cout << ans[0][i] << " ";
}

```