You are given N-1 subsets of \\{1,2,...,N\\}. Let the i-th set be E_i.

Let us choose two distinct elements u_i and v_i from each set E_i, and consider a graph T with N vertices and N-1 edges, whose vertex set is \\{1,2,..,N\\} and whose edge set is (u_1,v_1),(u_2,v_2),...,(u_{N-1},v_{N-1}). Determine if T can be a tree by properly deciding u_i and v_i. If it can, additionally find one instance of (u_1,v_1),(u_2,v_2),...,(u_{N-1},v_{N-1}) such that T is actually a tree.

Constraints

* 2 \leq N \leq 10^5
* E_i is a subset of \\{1,2,..,N\\}.
* |E_i| \geq 2
* The sum of |E_i| is at most 2 \times 10^5.

Input

Input is given from Standard Input in the following format:


N
c_1 w_{1,1} w_{1,2} ... w_{1,c_1}
:
c_{N-1} w_{N-1,1} w_{N-1,2} ... w_{N-1,c_{N-1}}


Here, c_i stands for the number of elements in E_i, and w_{i,1},...,w_{i,c_i} are the c_i elements in c_i. Here, 2 \leq c_i \leq N, 1 \leq w_{i,j} \leq N, and w_{i,j} \neq w_{i,k} (1 \leq j < k \leq c_i) hold.

Output

If T cannot be a tree, print `-1`; otherwise, print the choices of (u_i,v_i) that satisfy the condition, in the following format:


u_1 v_1
:
u_{N-1} v_{N-1}

Output

If T cannot be a tree, print `-1`; otherwise, print the choices of (u_i,v_i) that satisfy the condition, in the following format:


u_1 v_1
:
u_{N-1} v_{N-1}

Examples

Input

5
2 1 2
3 1 2 3
3 3 4 5
2 4 5


Output

1 2
1 3
3 4
4 5


Input

6
3 1 2 3
3 2 3 4
3 1 3 4
3 1 2 4
3 4 5 6


Output

-1


Input

10
5 1 2 3 4 5
5 2 3 4 5 6
5 3 4 5 6 7
5 4 5 6 7 8
5 5 6 7 8 9
5 6 7 8 9 10
5 7 8 9 10 1
5 8 9 10 1 2
5 9 10 1 2 3


Output

1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>

using namespace std;

#define LL long long
#define fgx cerr<<"--------------"<<endl;
#define dgx cerr<<"=============="<<endl;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN = 600010;
const int INF = 2147483600;

int N;

int Node[MAXN<<1],Next[MAXN<<1],Root[MAXN+1],C[MAXN<<1],cnt; 
inline void addedge(int u,int v,int c){
	Node[cnt]=v; Next[cnt]=Root[u]; C[cnt]=c; Root[u]=cnt++;
}
inline void insert(int u,int v,int c){
	addedge(u,v,c); addedge(v,u,0);
}
int S,T; int dis[MAXN+1];
inline bool BFS(){
	for(int i=S;i<=T;i++) dis[i]=0;
	queue<int> que; que.push(S); dis[S]=1;
	while(!que.empty()){
		int k=que.front();
		for(int x=Root[k];x!=-1;x=Next[x]){
			int v=Node[x]; 
			if(C[x]>0&&!dis[v]) dis[v]=dis[k]+1,que.push(v); 
		} que.pop();
	} 
	return (dis[T]?1:0);
}
inline int DFS(int k,int t){
	if(k==T) {return t;} int ret=0;
	for(int x=Root[k];x!=-1;x=Next[x]){
		int v=Node[x]; if(C[x]>0&&dis[v]==dis[k]+1){
			int d=DFS(v,min(t,C[x]));
			C[x]-=d; C[x^1]+=d; ret+=d; t-=d;
			if(!t) return ret;
		}
	} if(!ret) dis[k]=0;
	return ret;
}
int nod;
queue<int> que; int vis[MAXN+1],chs[MAXN+1],root;
vector<int> vec[MAXN+1]; int P[MAXN+1];
inline void Add(int k){
	++nod; for(int i=0,sz=vec[k].size();i<sz;i++){
		int j=vec[k][i];
		if(!vis[j]&&chs[j]!=k) vis[j]=k,que.push(j); 
	}
}

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	N=read(); T=2*N,S=0;
	memset(Root,-1,sizeof(Root));
	for(int i=1;i<N;i++){
		P[i]=read();
		insert(S,i,1);
		for(int j=1;j<=P[i];j++){
			int x=read();
			vec[x].push_back(i);
			insert(i,x+N-1,1);
		}
	}
	for(int i=1;i<=N;i++) insert(i+N-1,T,1);
	int sum=0; while(BFS()) sum+=DFS(S,INF);
	if(sum!=N-1){ puts("-1"); return 0; }
	int id=1; int now=0;
	for(int i=1;i<N;i++){
		now+=2;
		for(int j=1;j<=P[i];j++){
			if(!C[now]) chs[i]=Node[now]-(N-1);
			now+=2;
		}
	}
	while(now<cnt) {
		if(C[now]) {
			root=id; break;
		} now+=2; ++id;
	} 
	Add(root);
	while(!que.empty()){ int k=que.front();  Add(chs[k]); que.pop(); }
	if(nod!=N){ puts("-1"); return 0; }
	for(int i=1;i<N;i++) printf("%d %d\n",chs[i],vis[i]);
	return 0;
}


