#include<bits/stdc++.h>
#define rep(i,a,b) for (int i = (a); i <= (b); ++i)
#define drep(i,a,b) for (int i = (a); i >= (b); --i)
#define grep(i,u) for (int i = head[u],v = e[i].v; i; v = e[i = e[i].nxt].v)
#define il inline
#define LL long long
#define ULL unsigned LL
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
il LL read() {
	LL res = 0,f = 1; char ch = getchar(); while (!isdigit(ch)) {if (ch == '-') f = -f; ch = getchar();}
	while (isdigit(ch)) {res = res*10+ch-'0'; ch = getchar();} return res*f;
}
namespace qiqi {
	const int N = 1e5+5,INF = 0x3f3f3f3f; int n,ecnt,head[N],dis[N][3],a[N],m,d[N];
	struct Edge {int v,nxt;} e[N<<1]; il void add(int u,int v) {e[++ecnt] = (Edge){v,head[u]}; head[u] = ecnt;}
	int find(int u,int fa,int c) {int mx = u,x; grep(i,u) if (v != fa) {dis[v][c] = dis[u][c]+1; x = find(v,u,c); if (dis[x][c] > dis[mx][c]) mx = x;} return mx;}
	void dfs(int u,int fa) {int x = dis[u][1]+dis[u][2]; a[++m] = u; grep(i,u) if (v != fa && x == dis[v][1]+dis[v][2]) {dfs(v,u); return;}}
	void main() {
		n = read(); rep(i,2,n) {int u = read(),v = read(); add(u,v); add(v,u);} dfs(find(find(find(1,0,0),0,1),0,2),0);
		int x = dis[a[1]][1]+dis[a[1]][2],y; rep(u,1,n) {if ((y = dis[u][1]+dis[u][2]) > x+2) {puts("-1"); return;} if (y > x) grep(i,u) ++d[v];}
		rep(i,1,m/2) {x = a[i]; y = a[m-i+1]; if (d[x] != d[y]) {if (d[x] > d[y]) reverse(a+1,a+m+1); break;}}
		x = 1; rep(j,1,m) {y = x+d[a[j]]-1; rep(i,x,y) printf("%d ",i+1); printf("%d ",x); x = y+2;}
	}
}
int main() {
	qiqi::main(); 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>
#define rep(i,a,b) for (int i = (a); i <= (b); ++i)
#define drep(i,a,b) for (int i = (a); i >= (b); --i)
#define grep(i,u) for (int i = head[u],v = e[i].v; i; v = e[i = e[i].nxt].v)
#define il inline
#define LL long long
#define ULL unsigned LL
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
il LL read() {
	LL res = 0,f = 1; char ch = getchar(); while (!isdigit(ch)) {if (ch == '-') f = -f; ch = getchar();}
	while (isdigit(ch)) {res = res*10+ch-'0'; ch = getchar();} return res*f;
}
namespace qiqi {
	const int N = 1e5+5,INF = 0x3f3f3f3f; int n,ecnt,head[N],dis[N][3],a[N],m,d[N];
	struct Edge {int v,nxt;} e[N<<1]; il void add(int u,int v) {e[++ecnt] = (Edge){v,head[u]}; head[u] = ecnt;}
	int find(int u,int fa,int c) {int mx = u,x; grep(i,u) if (v != fa) {dis[v][c] = dis[u][c]+1; x = find(v,u,c); if (dis[x][c] > dis[mx][c]) mx = x;} return mx;}
	void dfs(int u,int fa) {int x = dis[u][1]+dis[u][2]; a[++m] = u; grep(i,u) if (v != fa && x == dis[v][1]+dis[v][2]) {dfs(v,u); return;}}
	void main() {
		n = read(); rep(i,2,n) {int u = read(),v = read(); add(u,v); add(v,u);} dfs(find(find(find(1,0,0),0,1),0,2),0);
		int x = dis[a[1]][1]+dis[a[1]][2],y; rep(u,1,n) {if ((y = dis[u][1]+dis[u][2]) > x+2) {puts("-1"); return;} if (y > x) grep(i,u) ++d[v];}
		rep(i,1,m/2) {x = a[i]; y = a[m-i+1]; if (d[x] != d[y]) {if (d[x] > d[y]) reverse(a+1,a+m+1); break;}}
		x = 1; rep(j,1,m) {y = x+d[a[j]]-1; rep(i,x,y) printf("%d ",i+1); printf("%d ",x); x = y+2;}
	}
}
int main() {
	qiqi::main(); return 0;
}
```