Problem: Cirno has prepared n arrays of length n each. Each array is a permutation of n integers from 1 to n. These arrays are special: for all 1 ≤ i ≤ n, if we take the i-th element of each array and form another array of length n with these elements, the resultant array is also a permutation of n integers from 1 to n. In the other words, if you put these n arrays under each other to form a matrix with n rows and n columns, this matrix is a [Latin square](https://en.wikipedia.org/wiki/Latin_square).

Afterwards, Cirno added additional n arrays, each array is a permutation of n integers from 1 to n. For all 1 ≤ i ≤ n, there exists at least one position 1 ≤ k ≤ n, such that for the i-th array and the (n + i)-th array, the k-th element of both arrays is the same. Notice that the arrays indexed from n + 1 to 2n don't have to form a Latin square. 

Also, Cirno made sure that for all 2n arrays, no two arrays are completely equal, i. e. for all pair of indices 1 ≤ i < j ≤ 2n, there exists at least one position 1 ≤ k ≤ n, such that the k-th elements of the i-th and j-th array are different.

Finally, Cirno arbitrarily changed the order of 2n arrays.

AquaMoon calls a subset of all 2n arrays of size n good if these arrays from a Latin square.

AquaMoon wants to know how many good subsets exist. Because this number may be particularly large, find it modulo 998 244 353. Also, she wants to find any good subset. Can you help her?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (5 ≤ n ≤ 500).

Then 2n lines followed. The i-th of these lines contains n integers, representing the i-th array.

It is guaranteed, that the sum of n over all test cases does not exceed 500.

Output

For each test case print two lines.

In the first line, print the number of good subsets by modulo 998 244 353.

In the second line, print n indices from 1 to 2n — indices of the n arrays that form a good subset (you can print them in any order). If there are several possible answers — print any of them.

Example

Input


3
7
1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6
1 2 3 4 5 7 6
1 3 4 5 6 7 2
1 4 5 6 7 3 2
1 5 6 7 4 2 3
1 6 7 5 2 3 4
1 7 6 2 3 4 5
1 7 2 3 4 5 6
5
4 5 1 2 3
3 5 2 4 1
1 2 3 4 5
5 2 4 1 3
3 4 5 1 2
2 3 4 5 1
1 3 5 2 4
4 1 3 5 2
2 4 1 3 5
5 1 2 3 4
6
2 3 4 5 6 1
3 1 2 6 4 5
6 1 2 3 4 5
5 6 1 3 2 4
4 3 6 5 2 1
5 6 1 2 3 4
4 5 6 1 2 3
3 4 5 6 1 2
1 2 3 4 5 6
2 5 4 1 6 3
3 2 5 4 1 6
1 4 3 6 5 2


Output


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

Note

In the first test case, the number of good subsets is 1. The only such subset is the set of arrays with indices 1, 2, 3, 4, 5, 6, 7.

In the second test case, the number of good subsets is 2. They are 1, 3, 5, 6, 10 or 2, 4, 7, 8, 9.

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

//======================
//
// Template OleschY
//
// Die M�glichkeiten der deutschen Grammatik k�nnen
// einen, wenn man sich darauf, was man ruhig, wenn
// man m�chte, sollte, einl�sst, �berraschen.
// - @Gedankenbalsam
//
//======================

//======================
//      Output/Debug
//======================

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "[";
    for (int i = 0; i < v.size(); ++i)
    {
        os << v[i];
        if (i != v.size() - 1)
            os << ", ";
    }
    os << "]\n";
    return os;
}

template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p)
{
    os << "[";
    os << p.first;
    os << ", ";
    os << p.second;
    os << "]";
    return os;
}

#ifndef ONLINE_JUDGE
#define debug(x) cout << (#x) << " is " << (x) << endl
#else
#define debug(...) 42
#endif

//======================
//      MODULE
//======================
struct ModInt;
vector<ModInt> inverse();

using ModType = long long; // set Type. Cann multiply ints if using long long.
struct ModInt
{
    static inline vector<ModInt> precalcInverse;
    static void precalc()
    {
        precalcInverse = inverse();
    }

    static inline ModType mod = 998244353; // Set this value
    ModType val;
    ModInt(ModType v = 0) : val(v % mod) { }
    static void setMod(ModType m)
    {
        mod = m;
        if(precalcInverse.size() > 0)
        {
            precalcInverse = inverse();
        }
    }
    ModInt &operator+=(ModInt a)
    {
        if ((val += a.val) >= mod) val -= mod;
        return *this;
    }
    ModInt &operator-=(ModInt a)
    {
        if ((val -= a.val) < 0) val += mod;
        return *this;
    }
    ModInt &operator*=(ModInt a)
    {
        val = (val * a.val) % mod;
        return *this;
    }
    ModInt &operator/=(ModInt a)
    {
        if(precalcInverse.size() > a.val)
        {
            *this *= precalcInverse[a.val];
            return *this;
        }
        ModType u = 1, v = a.val, s = 0, t = mod;
        while (v)
        {
            ModType q = t / v;
            swap(s -= u * q, u);
            swap(t -= v * q, v);
        }
        a.val = (s < 0 ? s + mod : s);
        val /= t;
        return (*this) *= a;
    }
    ModInt inv() const
    {
        return ModInt(1) /= (*this);
    }
    bool operator<(ModInt x) const
    {
        return val < x.val;
    }
};
ostream &operator<<(ostream &os, ModInt a)
{
    os << a.val;
    return os;
}
ModInt operator+(ModInt a, ModInt b)
{
    return a += b;
}
ModInt operator-(ModInt a, ModInt b)
{
    return a -= b;
}
ModInt operator*(ModInt a, ModInt b)
{
    return a *= b;
}
ModInt operator/(ModInt a, ModInt b)
{
    return a /= b;
}
ModInt pow(ModInt a, ModType e)
{
    ModInt x(1);
    for (; e > 0; e /= 2)
    {
        if (e % 2 == 1) x *= a;
        a *= a;
    }
    return x;
}
// compute inv[1], inv[2], ..., inv[mod-1] in O(n) time
vector<ModInt> inverse()
{
    ModType mod = ModInt::mod;
    vector<ModInt> inv(mod);
    inv[1].val = 1;
    for (ModType a = 2; a < mod; ++a)
        inv[a] = inv[mod % a] * ModInt(mod - mod / a);
    return inv;
}
ModInt stringToModInt(string s)
{
    ModType val = 0;
    for (int i = 0; i < s.size(); ++i)
        val = (val * 10 + (s[i] - '0')) % ModInt::mod;
    return ModInt(val);
}

//======================
// Solution Single testcase
//======================

int n;
vector<vector<int>> num; // array, column
vector<vector<int>> cnt; // column, number
vector<bool> chosen;
int nchosen;
vector<int> ans;
int anscomb;

void remrow(int row)
{
    if(chosen[row])
        return;
    chosen[row] = true;
    nchosen++;
    for(int col = 0; col < n; ++col)
    {
        cnt[col][num[row][col]]--;
    }
}


void choose(int row)
{
    ans.push_back(row);
    remrow(row);
    for(int row2 = 0; row2 < 2 * n; ++row2)
    {
        for(int col = 0; col < n; ++col)
        {
            if(num[row][col] == num[row2][col])
            {
                remrow(row2);
                break;
            }
        }
    }
}

void choose(int col, int val)
{
    for(int row = 0; row < 2 * n; row++)
    {
        if(chosen[row])
            continue;
        if(num[row][col] == val)
        {
            choose(row);
            break;
        }
    }

}


int solve()
{
    cin >> n;
    ans.clear();
    anscomb = 0;
    nchosen = 0;
    num.clear();
    cnt.clear();
    chosen.clear();
    num.resize(2 * n, vector<int>(n)); // array, column
    cnt.resize(n, vector<int>(n, 0)); // column, number
    chosen.resize(2 * n, false);
    for(int i = 0; i < 2 * n; ++i)
    {
        for(int j = 0; j < n; ++j)
        {
            cin >> num[i][j];
            num[i][j]--;
            cnt[j][num[i][j]]++;
        }
    }
    ModInt comb = 1;
    while(true)
    {
        bool found = false;;
        for(int col = 0; col < n; ++col)
        {
            for(int val = 0; val < n; ++val)
            {
                if(cnt[col][val] == 1)
                {
                    choose(col, val);
                    found = true;
                }
            }
        }
        if(nchosen == 2 * n)
        {
            break;
        }
        if(!found)
        {
            for(int row = 0; row < 2 * n; ++row)
            {
                if(!chosen[row])
                {
                    choose(row);
                    comb *= 2;
                    break;
                }
            }
        }
        if(nchosen == 2 * n)
        {
            break;
        }
    }

    debug(ans);
    cout << comb << "\n";
    for(int i = 0; i < n; i++)
    {
        cout << ans[i] + 1 << " ";
    }
    cout << "\n";
    return 0;
}

//======================
// Technical stuff
//======================

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    //freopen("a.out","w",stdout); // console output is better (in most cases)
#else
    // add i/o method of specific testing system
#endif
    int ntest = 1;
    cin >> ntest;
    for(int test = 0; test < ntest; ++test)
    {
        solve();
    }
    return 0;
}