Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 200500
queue<int> tp;
int n,m,a,head[N],cnt=1,id[N],fa[N],cur[N],dis[N],is[N],ct1;
vector<int> st2[N];
struct edge{int t,next,v;}ed[N*4];
void adde(int f,int t,int v){ed[++cnt]=(edge){t,head[f],v};head[f]=cnt;ed[++cnt]=(edge){f,head[t],0};head[t]=cnt;}
bool bfs(int s,int t)
{
	memset(dis,-1,sizeof(dis));
	memcpy(cur,head,sizeof(cur));
	dis[s]=0;
	queue<int> st;st.push(s);
	while(!st.empty())
	{
		int a=st.front();st.pop();
		for(int i=head[a];i;i=ed[i].next)
		if(ed[i].v&&dis[ed[i].t]==-1)
		{
			dis[ed[i].t]=dis[a]+1,st.push(ed[i].t);
			if(ed[i].t==t)return 1;
		}
	}
	return 0;
}
int dfs(int u,int t,int f)
{
	if(u==t||!f)return f;
	int as=0,tp;
	for(int &i=cur[u];i;i=ed[i].next)
	if(ed[i].v&&dis[ed[i].t]==dis[u]+1&&(tp=dfs(ed[i].t,t,min(ed[i].v,f))))
	{
		as+=tp,f-=tp;ed[i].v-=tp;ed[i^1].v+=tp;
		if(!f)return as;
	}
	return as;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%d",&m);
		for(int j=1;j<=m;j++)scanf("%d",&a),st2[a].push_back(i),adde(a,i+n,1);
	}
	for(int i=1;i<n;i++)adde(2*n,i+1,1),adde(i+n,2*n+1,1);
	int as=0;while(bfs(n*2,n*2+1))
	as+=dfs(n*2,n*2+1,n);
	if(as<n-1){printf("-1\n");return 0;}
	for(int i=1;i<n;i++)
	for(int j=head[i+n];j;j=ed[j].next)if(ed[j].v)id[i]=ed[j].t;
	queue<int> st;
	st.push(1);
	while(!st.empty())
	{
		int a=st.front();st.pop();
		ct1++;
		for(int j=0;j<st2[a].size();j++)if(!fa[id[st2[a][j]]])fa[id[st2[a][j]]]=a,st.push(id[st2[a][j]]);
	}
	if(ct1<n){printf("-1\n");return 0;}
	for(int i=1;i<n;i++)printf("%d %d\n",fa[id[i]],id[i]);
}