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.
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#include <unordered_set>
#include <unordered_map>
#define rep(x, s, t) for(llint x = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(llint x = (s); (x) >= (t); (x)--)
#define reps(x, s) for(llint x = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define outf(x) cout << fixed << setprecision(16) << (x) << endl
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

//const ll mod = 1000000007;
const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

vector<mint> fact, fact_inv;
void make_fact(int n){
	fact.resize(n+1), fact_inv.resize(n+1);
	fact[0] = mint(1); rep(i, 1, n) fact[i] = fact[i-1] * mint(i);
	fact_inv[n] = fact[n].inverse(); per(i, n-1, 0) fact_inv[i] = fact_inv[i+1] * mint(i+1);
}
mint comb(ll n, ll k){ if(n < 0 || k < 0 || n < k) return mint(0); return fact[n] * fact_inv[k] * fact_inv[n-k];}
mint perm(ll n, ll k){ return comb(n, k) * fact[k]; }

vector<int> prime;
void make_prime(int n){
	prime.resize(n+1);
	rep(i, 2, n){
		if(prime[i]) continue;
		for(int j = i; j <= n; j+=i) prime[j] = i;
	}
}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "YES" << endl; }
void no(){ cout << "NO" << endl; }
ll sgn(ll x){ if(x > 0) return 1; if(x < 0) return -1; return 0;}
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll lcm(ll a, ll b){return a/gcd(a, b)*b;}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}
string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;}
ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;}
template<typename T>
void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}

template<class S, class T> pair<S, T>& operator+=(pair<S,T> &s, const pair<S,T> &t){
	s.first += t.first, s.second += t.second;
	return s;
}
template<class S, class T> pair<S, T>& operator-=(pair<S,T> &s, const pair<S,T> &t){
	s.first -= t.first, s.second -= t.second;
	return s;
}
template<class S, class T> pair<S, T> operator+(const pair<S,T> &s, const pair<S,T> &t){
	return pair<S,T>(s.first+t.first, s.second+t.second);
}
template<class S, class T> pair<S, T> operator-(const pair<S,T> &s, const pair<S,T> &t){
	return pair<S,T>(s.first-t.first, s.second-t.second);
}
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++; if(itr != map_var.end()) os << ","; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, multiset<T>& set_var) {
	for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t){for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}cout << endl;}
void dump_func(){cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}

ll T;
ll n;
ll a[1005][505];
bool used[1005];
ll have[1005];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> T;
	while(T--){
		cin >> n;
		rep(i, 1, 2*n) rep(j, 1, n) cin >> a[i][j];
		
		mint ans = 1; ll rem = 2*n; vector<ll> avec;
		rep(i, 1, 2*n) used[i] = false;
		while(rem){
			ll p = 0;
			rep(j, 1, n){
				rep(k, 1, n) have[k] = 0;
				rep(k, 1, 2*n){
					if(used[k]) continue;
					if(have[a[k][j]] == 0) have[a[k][j]] = k;
					else have[a[k][j]] = -1;
				}
				rep(k, 1, n){
					if(have[k] > 0){
						p = have[k];
						goto end;
					}
				}
			}
			end:;
			if(p == 0){
				ans *= mint(2);
				rep(j, 1, 2*n){
					if(!used[j]){
						p = j;
						break;
					}
				}
			}
			avec.push_back(p);
			rep(j, 1, 2*n){
				if(used[j]) continue;
				rep(k, 1, n) if(a[j][k] == a[p][k]){
					used[j] = true;
					rem--;
					break;
				}
			}
		}
		outl(ans);
		outl(avec);
	}
	
	return 0;
}