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:
// --------------------------------------------------<TEMPLATE>--------------------------------------------------
// --------------------<optimizations>--------------------
/*
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
*/ 
// -------------------</optimizations>--------------------
 
// ---------------<Headers and namespaces>----------------
#include <bits/stdc++.h>
using namespace std;
 
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/
 
// ---------------</Headers and namespaces>---------------
 
// -----------------<Defines and typedefs>----------------
// typedef tree<int,null_type,less<int>,rb_tree_tag, 
// tree_order_statistics_node_update> indexed_set; // use less_equal for multiset
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the iterator to kth largest element.(0-based)
 
typedef long double LD;
typedef long long ll;
#define int ll
#define pb push_back
#define mp make_pair
#define REP(i,n) for (int i = 0; i < n; i++)
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define REPD(i,n) for (int i = n-1; i >= 0; i--)
#define FORD(i,a,b) for (int i = a; i >= b; i--)
#define remax(a,b) a = max(a,b)
#define remin(a,b) a = min(a,b)
#define all(v) v.begin(),v.end()
typedef map<int,int> mii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
#define F first
#define S second
#define PQ(type) priority_queue<type>
#define PQD(type) priority_queue<type,vector<type>,greater<type> >
#define WL(t) while(t --)
#define SZ(x) ((int)(x).size())
#define runtime() ((double)clock() / CLOCKS_PER_SEC)
#define inrange(i,a,b) ((i>=min(a,b)) && (i<=max(a,b)))
 
/*
// -----<SCANF>-----
#define sfi(x) scanf("%d",&x);
#define sfi2(x,y) scanf("%d%d",&x,&y);
#define sfi3(x,y,z) scanf("%d%d%d",&x,&y,&z);
 
#define sfl(x) scanf("%lld",&x);
#define sfl2(x,y) scanf("%lld%lld",&x,&y);
#define sfl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z);
#define sfl4(x,y,z,x1) scanf("%lld%lld%lld%lld",&x,&y,&z,&x1);
#define sfl5(x,y,z,x1,y1) scanf("%lld%lld%lld%lld%lld",&x,&y,&z,&x1,&y1);
#define sfl6(x,y,z,x1,y1,z1) scanf("%lld%lld%lld%lld%lld%lld",&x,&y,&z,&x1,&y1,&z1);
 
#define sfs(x) scanf("%s",x);
#define sfs2(x,y) scanf("%s%s",x,y);
#define sfs3(x,y,z) scanf("%s%s%s",x,y,z);
// ----</SCANF>-----
 
// ----<PRINTF>-----
#define pfi(x) printf("%d\n",x);
#define pfi2(x,y) printf("%d %d\n",x,y);
#define pfi3(x,y,z) printf("%d %d %d\n",x,y,z);
 
#define pfl(x) printf("%lld\n",x);
#define pfl2(x,y) printf("%lld %lld\n",x,y);
#define pfl3(x,y,z) printf("%lld %lld %lld\n",x,y,z);
 
#define pfs(x) printf("%s\n",x);
#define pfs2(x,y) printf("%s %s\n",x,y);
#define pfs3(x,y,z) printf("%s %s %s\n",x,y,z);
 
#define pwe(x) printf("%lld ",x); // print without end
// ----</PRINTF>----
*/
#define FLSH fflush(stdout)
#define fileIO(name) \
    freopen(name".in", "r", stdin); \
    freopen(name".out", "w", stdout);
#define PRECISION(x) cout << fixed << setprecision(x); 
#define FAST_IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
 
// ----------------</Defines and typedefs>----------------
 
// -------------------<Debugging stuff>-------------------
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
// ------------------</Debugging stuff>-------------------
 
// ------------------------<Consts>-----------------------
const int MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI=acos(-1);
 
#ifdef int
const int INF=1e16;
#else
const int INF=1e9;
#endif
 
const int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
// -----------------------</Consts>-----------------------

// -------------------------<RNG>-------------------------
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count()); 
#define SHUF(v) shuffle(all(v), RNG);
// Use mt19937_64 for 64 bit random numbers.
 
// ------------------------</RNG>-------------------------
 
// ----------------------<MATH>---------------------------
template<typename T> T gcd(T a, T b){return(b?__gcd(a,b):a);}
template<typename T> T lcm(T a, T b){return(a*(b/gcd(a,b)));}
int add(int a, int b, int c = MOD){int res=a+b;return(res>=c?res-c:res);}
int mod_neg(int a, int b, int c = MOD){int res;if(abs(a-b)<c)res=a-b;else res=(a-b)%c;return(res<0?res+c:res);}
int mul(int a, int b, int c = MOD){ll res=(ll)a*b;return(res>=c?res%c:res);}
int muln(int a, int b, int c = MOD){ll res=(ll)a*b;return ((res%c)+c)%c;}
ll mulmod(ll a,ll b, ll m = MOD){ll q = (ll)(((LD)a*(LD)b)/(LD)m);ll r=a*b-q*m;if(r>m)r%=m;if(r<0)r+=m;return r;}
template<typename T>T expo(T e, T n){T x=1,p=e;while(n){if(n&1)x=x*p;p=p*p;n>>=1;}return x;}
template<typename T>T power(T e, T n, T m = MOD){T x=1,p=e;while(n){if(n&1)x=mul(x,p,m);p=mul(p,p,m);n>>=1;}return x;}
template<typename T>T extended_euclid(T a, T b, T &x, T &y){T xx=0,yy=1;y=0;x=1;while(b){T q=a/b,t=b;b=a%b;a=t;\
t=xx;xx=x-q*xx;x=t;t=yy;yy=y-q*yy;y=t;}return a;}
template<typename T>T mod_inverse(T a, T n = MOD){T x,y,z=0;T d=extended_euclid(a,n,x,y);return(d>1?-1:mod_neg(x,z,n));}
 
const int FACSZ = 1; // Make sure to change this
 
int fact[FACSZ],ifact[FACSZ];
 
void precom(int c = MOD){
    fact[0] = 1;
    FOR(i,1,FACSZ) fact[i] = mul(fact[i-1],i,c);
    ifact[FACSZ-1] = mod_inverse(fact[FACSZ-1],c);
    REPD(i,FACSZ-1){
        ifact[i] = mul(i+1,ifact[i+1],c);
    }
}
 
int ncr(int n,int r,int c = MOD){
    if(r > n) return 0;
    return mul(mul(ifact[r],ifact[n-r],c),fact[n],c);
} 

// ----------------------</MATH>--------------------------
// --------------------------------------------------</TEMPLATE>--------------------------------------------------
 
void solvethetestcase();
 
signed main(){
    // comment when using scanf,printf
    FAST_IO
    int t = 1;
    // (UNCOMMENT FOR MULTIPLE TEST CASES)
    cin >> t;
    FOR(testcase,1,t+1){
        // (UNCOMMENT FOR CODEJAM)
        // printf("Case #%lld: ",testcase); 
        solvethetestcase();
    }
}   
 
int n;
int a[1005][505];
int vis[1005];
int cnt[505][505];
 
void solvethetestcase(){
    cin >> n;
    int ans = 1;
    vi res;
    REP(i,2*n){
        vis[i] = 0;
        REP(j,n){
            cin >> a[i][j];
            a[i][j]--;
        }
    }
    while(SZ(res) < n){
        REP(i,n){
            REP(j,n){
                cnt[i][j] = 0;
            }
        }
        REP(i,2*n){
            if(vis[i]) continue;
            REP(j,n){
                cnt[j][a[i][j]]++;
            }
        }
        int cur = -1;
        REP(i,2*n){
            if(vis[i]) continue;
            REP(j,n){
                if(cnt[j][a[i][j]] == 1){
                    cur = i;
                    break;
                }
            }
            if(cur != -1) break;
        }
        if(cur != -1){
            res.pb(cur);
            REP(i,2*n){
                if(vis[i]) continue;
                REP(j,n){
                    if(a[i][j] == a[cur][j]){
                        vis[i] = 1;
                        break;
                    }
                }
            }
        }
        else{
            ans = (ans*2)%FMOD;
            REP(i,2*n){
                if(!vis[i]){
                    cur = i;
                    break;
                }
            }
            res.pb(cur);
            REP(i,2*n){
                if(vis[i]) continue;
                REP(j,n){
                    if(a[i][j] == a[cur][j]){
                        vis[i] = 1;
                        break;
                    }
                }
            }
        }
    }
    cout <<  ans << "\n";
    for(auto x:res) cout << x+1 << " ";
    cout << "\n";
}