Problem: Three men, A, B and C, are eating sushi together. Initially, there are N pieces of sushi, numbered 1 through N. Here, N is a multiple of 3.

Each of the three has likes and dislikes in sushi. A's preference is represented by (a_1,\ ...,\ a_N), a permutation of integers from 1 to N. For each i (1 \leq i \leq N), A's i-th favorite sushi is Sushi a_i. Similarly, B's and C's preferences are represented by (b_1,\ ...,\ b_N) and (c_1,\ ...,\ c_N), permutations of integers from 1 to N.

The three repeats the following action until all pieces of sushi are consumed or a fight brakes out (described later):

* Each of the three A, B and C finds his most favorite piece of sushi among the remaining pieces. Let these pieces be Sushi x, y and z, respectively. If x, y and z are all different, A, B and C eats Sushi x, y and z, respectively. Otherwise, a fight brakes out.



You are given A's and B's preferences, (a_1,\ ...,\ a_N) and (b_1,\ ...,\ b_N). How many preferences of C, (c_1,\ ...,\ c_N), leads to all the pieces of sushi being consumed without a fight? Find the count modulo 10^9+7.

Constraints

* 3 \leq N \leq 399
* N is a multiple of 3.
* (a_1,\ ...,\ a_N) and (b_1,\ ...,\ b_N) are permutations of integers from 1 to N.

Input

Input is given from Standard Input in the following format:


N
a_1 ... a_N
b_1 ... b_N


Output

Print the number of the preferences of C that leads to all the pieces of sushi being consumed without a fight, modulo 10^9+7.

Examples

Input

3
1 2 3
2 3 1


Output

2


Input

3
1 2 3
1 2 3


Output

0


Input

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


Output

80


Input

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


Output

160


Input

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


Output

33600

Solution (in CPP):

Solution:
#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;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define Return(ans) { cout << (ans) << endl; return 0; }
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using Pdi = pair<double, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod

template<long long MOD>
struct ModInt{

    using ll = long long;
    ll val;

    void setval(ll v) { val = v % MOD; };
    ModInt(): val(0) {}
    ModInt(ll v) { setval(v); };

    ModInt operator+(const ModInt &x) const { return ModInt(val + x.val); }
    ModInt operator-(const ModInt &x) const { return ModInt(val - x.val + MOD); }
    ModInt operator*(const ModInt &x) const { return ModInt(val * x.val); }
    ModInt operator/(const ModInt &x) const { return *this * x.inv(); }
    ModInt operator-() const { return ModInt(MOD - val); }
    ModInt operator+=(const ModInt &x) { return *this = *this + x; }
    ModInt operator-=(const ModInt &x) { return *this = *this - x; }
    ModInt operator*=(const ModInt &x) { return *this = *this * x; }
    ModInt operator/=(const ModInt &x) { return *this = *this / x; }

    friend ostream& operator<<(ostream &os, const ModInt &x) { os << x.val; return os; }
    friend istream& operator>>(istream &is, ModInt &x) { is >> x.val; x.val = (x.val % MOD + MOD) % MOD; return is; }

    ModInt pow(ll n) const {
        ModInt a = 1;
        if(n == 0) return a;
        int i0 = 64 - __builtin_clzll(n);
        for(int i = i0 - 1; i >= 0; i--){
            a = a * a;
            if((n >> i) & 1) a *= (*this); 
        }
        return a;
    }
    ModInt inv() const { return this->pow(MOD - 2); }
};

using mint = ModInt<mod>; mint pow(mint x, long long n) { return x.pow(n); }
//using mint = double; //for debug
using mvec = vector<mint>;
using mmat = vector<mvec>;

struct Combination{

    vector<mint> fact, invfact;

    Combination(int N){
        fact = vector<mint>({mint(1)});
        invfact = vector<mint>({mint(1)});
        fact_initialize(N);
    }

    void fact_initialize(int N){
        int i0 = fact.size();
        if(i0 >= N + 1) return;
        fact.resize(N + 1);
        invfact.resize(N + 1);
        for(int i = i0; i <= N; i++) fact[i] = fact[i - 1] * i;
        invfact[N] = (mint)1 / fact[N];
        for(int i = N - 1; i >= i0; i--) invfact[i] = invfact[i + 1] * (i + 1); 
    }

    mint nCr(int n, int r){
        if(n < 0 || r < 0 || r > n) return mint(0);
        if(fact.size() < n + 1) fact_initialize(n);
        return fact[n] * invfact[r] * invfact[n - r];
    }

    mint nPr(int n, int r){
        if(n < 0 || r < 0 || r > n) return mint(0);
        if(fact.size() < n + 1) fact_initialize(n);
        return fact[n] * invfact[n - r];
    }

    mint Catalan(int n){
        if(n < 0) return 0;
        else if(n == 0) return 1;
        if(fact.size() < 2 * n + 1) fact_initialize(2 * n);
        return fact[2 * n] * invfact[n + 1] * invfact[n];
    }

};

signed main(){

    int N; cin >> N;
    vec a(N), b(N); cin >> a >> b;
    vec ord_a(N), ord_b(N);
    REP(i, N){
        a[i]--;
        b[i]--;
        ord_a[a[i]] = i;
        ord_b[b[i]] = i;
    }
    if(a[0] == b[0]){
        Out(0);
        return 0;
    }

    mat cnt(N, vec(N, 0));
    REP(x, N) REP(y, N){
        bools used(N, false);
        REP(i, x + 1) used[a[i]] = true;
        REP(i, y + 1) used[b[i]] = true;
        REP(i, N) if(used[i]) cnt[x][y]++;
    }
    //mdebug(cnt);

    vector<bools> valid(N, bools(N, true));
    REP(i, N) REP(j, N){
        if(ord_b[a[i]] <= j || ord_a[b[j]] <= i) valid[i][j] = false;
    }
    //mdebug(valid);

    Combination C(3 * N);
    int n = N / 3;

    vector<mmat> dp(n, mmat(N, mvec(N, 0)));
    REP(i, N) REP(j, N) if(valid[i][j]){
        int r = cnt[i][j] - 2 * n;
        if(r >= 0) dp[n - 1][i][j] = n - r;
    }

    IREP(t, n - 1){
        IREP(i, N) IREP(j, N - 1) dp[t + 1][i][j] += dp[t + 1][i][j + 1];
        IREP(j, N) IREP(i, N - 1) dp[t + 1][i][j] += dp[t + 1][i + 1][j];

        REP(i, N - 1) REP(j, N - 1) if(valid[i][j]){
            int r = cnt[i][j] - 2 * (t + 1);
            if(r >= 0 && r <= t + 1) dp[t][i][j] = dp[t + 1][i + 1][j + 1] * (t + 1 - r);
        }
    }

    mint ans = dp[0][0][0];

    mmat dq(n + 1, mvec(2 * n + 1, 0));
    dq[1][0] = 1;
    FOR(i, 1, n){
        int t = 2 * i;
        REP(j, t + 1) REP(k, j + 1){
            dq[i + 1][j] += dq[i][k] * C.nPr(t - k, j - k);
        }
    }
    mint c = 0;
    REP(j, 2 * n + 1) c += dq[n][j] * C.fact[2 * n - j];
    ans *= c;
    Out(ans);

    return 0;
}