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>
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template<class T> using pq = priority_queue<T>;
template<class T> using pqg = priority_queue<T, vector<T>, greater<T>>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a,x) for (auto& a : x)
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define ins insert

template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#define dbg(x...) cerr <<__func__<<":"<<__LINE__<<" [" << #x << "] = ["; _print(x); cerr <<  endl;


const int MOD = 998244353;
const char nl = '\n';
const int MX = 1001; 

template<int SZ> struct DSU {
    int parent[SZ], size[SZ];

    DSU() {
        F0R(i, SZ) parent[i] = i, size[i] = 0;
    }

    int get(int x) {
        if (parent[x] != x) parent[x] = get(parent[x]);
        return parent[x];
    }

    void unify(int x, int y) {
        x = get(x); y = get(y);
        if (x == y) return;
        if (size[x] < size[y]) swap(x, y);
        if (size[x] == size[y]) size[x]++;
        parent[y] = x;

    }
};

struct mi {
	ll v; explicit operator ll() const { return v; }
	mi() { v = 0; }
	mi(ll _v) { 
		v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
		if (v < 0) v += MOD;
	}
	friend bool operator==(const mi& a, const mi& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mi& a, const mi& b) { 
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { 
		return a.v < b.v; }
   
	mi& operator+=(const mi& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mi& operator-=(const mi& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mi& operator*=(const mi& m) { 
		v = v*m.v%MOD; return *this; }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mi operator-() const { return mi(-v); }
	mi& operator++() { return *this += 1; }
	mi& operator--() { return *this -= 1; }
    mi operator++(int) { mi temp; temp.v = v++; return temp; }
    mi operator--(int) { mi temp; temp.v = v--; return temp; }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
    friend ostream& operator<<(ostream& os, const mi& m) {
        os << m.v; return os;
    }
    friend istream& operator>>(istream& is, mi& m) {
        ll x; is >> x;
        m.v = x;
        return is;
    }
};


typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

void __print(mi X) {
    cout << X.v;
}

int N;
int A[MX][MX];
set<int> pos[MX][MX];
queue<pi> q;
set<int> rr;

void remove(int X) {
    rr.erase(X);
    F0R(i, N) {
        pos[i][A[X][i]].erase(X);
        if (sz(pos[i][A[X][i]]) == 1) {
            q.push({i, A[X][i]});
        }
    } 
}


void solve() {
    rr.clear();
    cin >> N;
    F0R(i, N) F0R(j, N) pos[i][j].clear();
    F0R(i,2* N) {
        rr.ins(i);
        F0R(j, N) {
            cin >> A[i][j]; A[i][j]--;
            pos[j][A[i][j]].ins(i);
        }
    }
    F0R(i, N) {
        F0R(j, N) {
           if (sz(pos[i][j]) == 1) {
               q.push({i, j});
           }
        }
    }
    vi ans;
    while (!q.empty()) {
        pi p = q.front(); q.pop();
        if (!sz(pos[p.f][p.s])) continue;
        int cur = *pos[p.f][p.s].begin();
        remove(cur);
        ans.pb(cur);
        F0R(i, N) {
            while (sz(pos[i][A[cur][i]])) {
                int X = *pos[i][A[cur][i]].begin();
                remove(X);
            }
        }
    }
    
    DSU<MX> dsu;
    vector<set<int>> graph(2*N);
    int nc = 0;
    trav(a, rr) {
        F0R(i, N) {
            trav(b, pos[i][A[a][i]]) {
                if (a != b) {
                    //dbg(a, b);
                    dsu.unify(a, b);
                    graph[a].ins(b);
                }
            }
        }
    }
    trav(a, rr) {
        if (dsu.get(a) == a) nc++;
    }
    cout << pow(mi(2), nc) << nl;
    int dist[2*N]; F0R(i, 2*N) dist[i] = -1;
    trav(a, rr) {
        if (dist[a] != -1) continue;
        dist[a] = 0;
        queue<int> qq; qq.push(a);
        while (!qq.empty()) {
            int v = qq.front(); qq.pop();
            //dbg(v, dist[v]);
            if (dist[v] % 2 == 0) ans.pb(v);
            trav(b, graph[v]) {
                if (dist[b] == -1) {
                    dist[b] = dist[v] + 1;
                    qq.push(b);
                }
            }
        }
    }
    F0R(i, sz(ans)) {
        cout << ans[i]+1 << " ";
    }
    cout << nl;

}
 
int main() {
    cin.tie(0)->sync_with_stdio(0); 
    cin.exceptions(cin.failbit);

    int T = 1;
    cin >> T;
    while(T--) {
        solve();
    }

	return 0;
}