Firetrucks Are Red

Lily is fascinated by numbers. She believes the whole world revolves around them, and that everything is connected by numbers. Her friends, Alice, Bob, Charlie and Diane, are not convinced. But she gives them an example:

Alice lives in house number 25 on her street, but that is exactly Bob’s age. Bob is born on June 4th, and Charlie was his parents’ fourth child. Finally, Diane has five fingers on her left hand, which happens to be the same as the number of toes that Bob has on his right foot!

This shows that her friends are all connected—either directly or indirectly—by numbers. But she still has to convince her family as well as her coworkers.

Given a group of nn individuals, and a set of numbers that describe each individual, help Lily come up with a proof that shows that everyone in this group is either directly or indirectly connected by numbers, or determine that this is not possible.

Input

The input consists of:

• One line with an integer nn (2≤n≤2⋅1052≤n≤2⋅105), the number of individuals in the group. The individuals are numbered from 11 to nn.

• nn lines, describing the individuals in the group.

  The iith such line starts with an integer mimi (1≤mi≤2⋅1051≤mi≤2⋅105), the number of numbers that describe individual ii.

  The remainder of the line has mimi distinct integers di,1,…,di,midi,1,…,di,mi (1≤di,j≤1091≤di,j≤109 for each jj), the set of numbers that describe individual ii.

It is guaranteed that the sum over all mimi is at most 2⋅1052⋅105.

Output

Output a proof in the form of n−1n−1 lines, each of which contains three integers pp, qq and rr, where pp and qq are distinct individuals that are both described by the number rr. Using only these relations, it must be possible to show that any pair of individuals in the group are connected either directly or indirectly.

If no such proof exists, output “impossible”. If there are multiple proofs, you may output any one of them.

Sample Input 1	Sample Output 1
6 2 17 10 1 5 2 10 22 3 17 22 9 2 17 8 3 9 22 16	impossible

Sample Input 2	Sample Output 2
6 2 17 10 2 5 10 2 10 22 3 17 22 9 2 17 8 3 9 22 16	1 3 10 2 3 10 3 4 22 4 5 17 4 6 9

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 998244353;
const int N = 2e5+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m*n / gcd(m, n);
}

int f[200010];
int find(int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
}
bool add(int x, int y) {
    int fx = find(x), fy = find(y);
    if (fx != fy) {
        f[fx] = fy;
        return true;
    }
    return false;
}
int main() {
    int n;
    scanf("%d", &n);
    map<int, vector<int>> m;
    for (int i = 1; i <= n; i++)
    {
        f[i] = i;
        int j, k;
        scanf("%d", &j);
        while (j--) {
            scanf("%d", &k);
            m[k].push_back(i);
        }
    }
    vector<pair<PII, int>> ans;
    for (auto i : m) {
        for (int j = 1; j < i.second.size(); j++)
            if (add(i.second[j - 1], i.second[j]))
                ans.push_back({ { i.second[j - 1], i.second[j] }, i.first });
    }
    if (ans.size() != n - 1)
        printf("impossible\n");
    else
        for (auto i : ans)
            printf("%d %d %d\n", i.first.first, i.first.second, i.second);
}