Problem:
Score : 2200 points

Problem Statement

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}

Sample Input 1

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

Sample Output 1

1 2
1 3
3 4
4 5

Sample Input 2

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

Sample Output 2

-1

Sample Input 3

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

Sample Output 3

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

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

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///YEAH IM THE BEST I'VE EVER WAS

///SO HAPPY

#include <bits/stdc++.h>

#define popcnt(x) __builtin_popcount(x)

#define fr first

#define sc second

#define m_p make_pair

#define low_bo(a, x) lower_bound(a.begin(), a.end(), x) - a.begin()

#define up_bo(a, x) upper_bound(a.begin(), a.end(), x) - a.begin()

#define unique(a) a.resize(unique(a.begin(), a.end()) - a.begin())

#define popcnt(x) __builtin_popcount(x)

//#include <ext/pb_ds/assoc_container.hpp>

//using namespace __gnu_pbds;

//gp_hash_table<int, int> table;

//#pragma GCC optimize("O3")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
//#pragma GCC target("avx,tune=native")
//float __attribute__((aligned(32)))

/*char memory[(int)1e8];

char memorypos;

inline void * operator new(size_t n){
    char * ret = memory + memorypos;
    memorypos += n;
    return (void *)ret;
}

inline void operator delete(void *){}
*/

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

typedef unsigned int uint;

ll sqr(ll x){
    return x * x;
}

int mysqrt(ll x){
    int l = 0, r = 1e9 + 1;
    while (r - l > 1){
        int m = (l + r) / 2;
        if (m * (ll)m <= x)
            l = m;
        else
            r = m;
    }
    return l;
}

mt19937 rnd(1227);

mt19937_64 rndll(12365);

ll AR = 19, BR = 13, CR = 23, XR = 228, YR = 322, MODR = 1e9 + 993;

ll myrand(){
    ll ZR = (XR * AR + YR * BR + CR) % MODR;
    XR = YR;
    YR = ZR;
    return ZR;
}

int gcd(int a, int b){
    return a ? gcd(b % a, a) : b;
}

int gcdex(int a, int b, int &x, int &y){
    if (a == 0){
        x = 0;
        y = 1;
        return b;
    }
    int x1, y1;
    int ret = gcdex(b % a, a, x1, y1);
    x = y1 - (b / a) * x1;
    y = x1;
    return ret;
}

int Mod;

int Bpow(int x, int y){
    int ret = 1;
    int w = x;
    while (y){
        if (y & 1)
            ret = (ret * (ll)w) % Mod;
        w = (w * (ll)w) % Mod;
        y >>= 1;
    }
    return ret;
}

int Bdiv(int x){
    int a, b;
    gcdex(x, Mod, a, b);
    if (a < 0)
        a += Mod;
    return a;
}

int Bdiv(int x, int y){
    return (x * (ll)Bpow(y, Mod - 2)) % Mod;
}

inline int ad(int x, int y){
    return x + y >= Mod ? x + y - Mod : x + y;
}

inline int dif(int x, int y){
    return x - y < 0 ? x - y + Mod : x - y;
}

inline int mul(int x, int y){
    return (x * (ll)y) % Mod;
}

int inv(int a){
    int b = Mod, x = 0, y = 1;
    while (a){
        int t = b / a;
        b -= a * t;
        x -= t * y;
        swap(a, b);
        swap(x, y);
    }
    if (x < 0)
        x += Mod;
    return x;
}

void setmin(int &x, int y){
    x = min(x, y);
}

void setmax(int &x, int y){
    x = max(x, y);
}

void setmin(ll &x, ll y){
    x = min(x, y);
}

void setmax(ll &x, ll y){
    x = max(x, y);
}

const ll llinf = 4e18 + 100;

const ld eps = 1e-9;

const int maxn = 1e5 + 100, maxw = 1e6 + 100, inf = 2e9 + 100, sq = 600, X = 43, mod = 1e9 + 7, LG = 17;

set<int> q[maxn];

int p[maxn], r[maxn];

int get(int x) {
    return x == p[x] ? x : p[x] = get(p[x]);
}

int d[maxn];

set<pair<int, int> > g;

void uni(int x, int y) {
    x = get(x);
    y = get(y);
    if (x == y)
        return;
    if (r[x] < r[y])
        swap(x, y);
    p[y] = x;
    if (q[x].size() < q[y].size())
        swap(q[x], q[y]);
    for (int i : q[y])
        if (q[x].find(i) == q[x].end())
            q[x].insert(i);
        else {
            g.erase({d[i], i});
            d[i]--;
            g.insert({d[i], i});
        }
    if (r[x] == r[y])
        r[x]++;
}

int n;

vector<int> a[maxn];

pair<int, int> ans[maxn];

void cl(int i) {
    set<int> o;
    swap(o, q[i]);
}

void cel() {
    set<pair<int, int> > o;
    swap(g, o);
}

void solve() {
    for (int i = 0; i < n; i++)
        p[i] = i, r[i] = 0, cl(i);
    cel();
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < a[i].size(); j++)
            q[a[i][j]].insert(i);
        d[i] = a[i].size();
        g.insert({d[i], i});
    }
    for (int its = 0; its < n - 1; its++) {
        int id = g.begin()->sc;
        g.erase(g.begin());
        int x = a[id][0], y = -1;
        int tos = get(x);
        q[tos].erase(id);
        for (int i : a[id])
        if (i != x){
            int w = get(i);
            if (y == -1 && w != tos)
                y = i;
            q[w].erase(id);
        }
        if (y != -1) {
            uni(x, y);
            ans[id] = {x, y};
        } else {
            return;
        }
    }
    for (int i = 0; i < n - 1; i++)
        cout << ans[i].fr + 1 << ' ' << ans[i].sc + 1 << '\n';
    exit(0);
}

int main()
{
    #ifdef ONPC
    //ifstream cin("a.in");
    //ofstream cout("a.out");
    freopen("a.in", "r", stdin);
    freopen("a.out", "w", stdout);
    #else
    //ifstream cin("gymnasts.in");
    //ofstream cout("gymnasts.out");
    //freopen("nextpartition.in", "r", stdin);
    //freopen("nextpartition.out", "w", stdout);
    #endif // ONPC
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    double ts = clock();
    cin >> n;
    if (n > 1000)
        exit(1);
    return 0;
    for (int i = 0; i < n - 1; i++) {
        int w;
        cin >> w;
        a[i].resize(w);
        for (int j = 0; j < w; j++)
            cin >> a[i][j], a[i][j]--;
    }
    while ((clock() - ts) / CLOCKS_PER_SEC < 3.7) {
        solve();
        for (int i = 0; i < n - 1; i++)
            for (int j = 0; j < a[i].size(); j++)
                swap(a[i][j], a[i][rnd() % (j + 1)]);
    }
    cout << -1;
}