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<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<cstring>
#include<climits>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
inline int read() {
	bool f=0;int x=0;char c=getchar();
	while(c<'0'||'9'<c){if(c==EOF)exit(0);if(c=='-')f=1;c=getchar();}
	while('0'<=c&&c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return !f?x:-x;
}
#define MAXN 200000
#define MAXM 500000
#define INF 0x3f3f3f3f
struct Edge{
	int nxt,v,cap;
}edge[2*MAXM+5];
int ecnt,head[MAXN+5],cur[MAXN+5];
void Init(){
	ecnt=-1,memset(head,-1,sizeof(head));
	return ;
}
void Addedge(int u,int v,int cap){
	//printf("%d %d %d\n",u,v,cap);
	edge[++ecnt]=(Edge){head[u],v,cap},head[u]=ecnt;
	edge[++ecnt]=(Edge){head[v],u,0},head[v]=ecnt;
	return ;
}
int N,S,T;
int dep[MAXN+5];
bool BFS(){
	queue<int> Q;
	for(int i=0;i<=N;i++)
		dep[i]=INF;
	dep[S]=0,Q.push(S);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int i=head[u];~i;i=edge[i].nxt){
			int v=edge[i].v,cap=edge[i].cap;
			if(dep[v]==INF&&cap)
				dep[v]=dep[u]+1,Q.push(v);
		}
	}
	return dep[T]<INF;
}
int DFS(int u,int aug){
	if(u==T) return aug;
	int flow=0,f;
	for(int &i=cur[u];~i;i=edge[i].nxt){
		int v=edge[i].v,cap=edge[i].cap;
		if(dep[v]==dep[u]+1&&cap&&(f=DFS(v,min(aug,cap)))){
			aug-=f,flow+=f;
			edge[i].cap-=f,edge[i^1].cap+=f;
			if(!aug) break;
		}
	}
	return flow;
}
int Dinic(){
	int Max_Flow=0;
	while(BFS())
		memcpy(cur,head,sizeof(head)),Max_Flow+=DFS(S,INF);
	return Max_Flow;
}
queue<int> Q;
int ma[MAXN+5];
bool vis[MAXN+5];
vector<int> G[MAXN+5];
int cho1[MAXN+5],cho2[MAXN+5];
int main(){
	Init();
	int n=read();
	N=2*n+2,S=2*n+1,T=2*n+2;
	for(int i=1;i<n;i++)
		Addedge(S,i,1);
	for(int i=2;i<=n;i++)	
		Addedge(i+n-1,T,1);
	for(int i=1;i<n;i++){
		int c=read();
		for(int j=1;j<=c;j++){
			int w=read();
			G[w].push_back(i);
			if(w!=1)
				Addedge(i,w+n-1,1);
		}
	}
	int ans=Dinic();
	if(ans!=n-1)
		puts("-1"),exit(0);
	for(int s=1;s<n;s++)
		for(int i=head[s];~i;i=edge[i].nxt)
			if(!edge[i].cap)
				ma[s]=edge[i].v-(n-1);
	Q.push(1);
	int cnt=0;
	while(!Q.empty()){
		cnt++;
		int u=Q.front();Q.pop();
		for(int i=0;i<(int)G[u].size();i++){
			int s=G[u][i];
			if(vis[s]) continue;
			vis[s]=1,Q.push(ma[s]);
			cho1[s]=u,cho2[s]=ma[s];
		}
	}
	if(cnt!=n)
		puts("-1"),exit(0);
	for(int i=1;i<n;i++)
		printf("%d %d\n",cho1[i],cho2[i]);
	return 0;
}
//