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:
// {{{ by shik
#include <bits/stdc++.h>
#include <unistd.h>
#pragma GCC diagnostic ignored "-Wunused-result"
#pragma GCC diagnostic ignored "-Wunused-function"
#define SZ(x) ((int)(x).size())
#define ALL(x) begin(x), end(x)
#define REP(i, n) for (int i = 0; i < int(n); i++)
#define REP1(i, a, b) for (int i = (a); i <= int(b); i++)
#define MP make_pair
#define PB push_back
using namespace std;
typedef int64_t LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

namespace { namespace shik {

template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%" SCNd64, &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }

template<class T> void _W(const T &x) { cout << x; }
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template<class T> void _W(const vector<T> &x) { for (auto i = x.begin(); i != x.end(); _W(*i++)) if (i != x.cbegin()) putchar(' '); }
void W() {}
template<class T, class... U> void W(const T &head, const U &... tail) { _W(head); putchar(sizeof...(tail) ? ' ' : '\n'); W(tail...); }

#ifdef SHIK
#include "dump.hpp"
#else
#define dump(...)
#endif

template<class T, class F = less<T>> void sort_uniq(vector<T> &v, F f = F()) { sort(begin(v), end(v), f); v.resize(unique(begin(v), end(v)) - begin(v)); }
template<class T> inline T bit(T x, int i) { return (x >> i) & 1; }
template<class T> inline bool chkmax(T &a, const T &b) { return b > a ? a = b, true : false; }
template<class T> inline bool chkmin(T &a, const T &b) { return b < a ? a = b, true : false; }
template<class T> using MaxHeap = priority_queue<T>;
template<class T> using MinHeap = priority_queue<T, vector<T>, greater<T>>;

// }}}

// {{{ ModInt
template<int MOD>
struct ModInt {
    int x;
    ModInt() : x(0) {}
    template<class T, typename enable_if<is_integral<T>::value>::type* = nullptr>
    ModInt(T _x) {
        x = _x % MOD;
        if (x < 0) x += MOD;
    }
    ModInt operator-() const { return {x == 0 ? 0 : MOD-x}; }
    ModInt& operator+=(ModInt rhs) {
        x += rhs.x;
        if (x >= MOD) x -= MOD;
        return *this;
    }
    ModInt& operator-=(ModInt rhs) {
        x -= rhs.x;
        if (x < 0) x += MOD;
        return *this;
    }
    ModInt& operator*=(ModInt rhs) {
        x = (long long)x * rhs.x % MOD;
        return *this;
    }
    ModInt& operator/=(ModInt rhs) {
        return *this *= rhs.inv();
    }
    ModInt operator+(ModInt rhs) const { return ModInt(*this) += rhs; }
    ModInt operator-(ModInt rhs) const { return ModInt(*this) -= rhs; }
    ModInt operator*(ModInt rhs) const { return ModInt(*this) *= rhs; }
    ModInt operator/(ModInt rhs) const { return ModInt(*this) /= rhs; }
    ModInt inv() const {
        // should work for non-prime MOD if gcd(x, MOD) = 1
        int a = x, b = MOD, u = 1, v = 0;
        while (b != 0) {
            int t = a / b;
            a -= t * b;
            u -= t * v;
            swap(a, b);
            swap(u, v);
        }
        return u;
    }
    ModInt pow(long long e) {
        ModInt r = 1, p = *this;
        while (e) {
            if (e & 1) r *= p;
            p *= p;
            e >>= 1;
        }
        return r;
    }
    bool operator==(ModInt rhs) { return x == rhs.x; }
    bool operator!=(ModInt rhs) { return x != rhs.x; }
    bool operator<(ModInt rhs) { return x < rhs.x; }
    bool operator<=(ModInt rhs) { return x <= rhs.x; }
    bool operator>(ModInt rhs) { return x > rhs.x; }
    bool operator>=(ModInt rhs) { return x >= rhs.x; }
    friend ostream& operator<<(ostream &os, ModInt i) { return os << i.x; }
};
// }}}

const int MOD = 1e9 + 7;
using mint = ModInt<MOD>;

const int N=410;
int n,a[N],b[N],cnt[N][N];
bitset<N> has[N][N];
mint dp[N][N][N/3][3];
void main() {
    R(n);
    REP1(i,1,n) R(a[i]);
    REP1(i,1,n) R(b[i]);
    REP1(i,1,n) {
        has[i][0]=has[i-1][0];
        has[i][0][a[i]]=1;
        REP1(j,1,n) {
            has[i][j]=has[i][j-1];
            has[i][j][b[j]]=1;
        }
    }
    REP1(i,1,n) REP1(j,1,n) cnt[i][j]=has[i][j].count();
    dp[1][1][0][0]=1;
    mint ans=0;
    REP1(i,1,n+1) REP1(j,1,n+1) REP1(k,0,n/3) REP(l,3) {
        mint me=dp[i][j][k][l];
        if ( me.x==0 ) continue;
        // dump(i,j,k,me);
        if ( i>n || j>n ) {
            if ( k==0 && l==0 ) ans+=me;
            continue;
        }
        if ( l==0 ) {
            if ( has[i-1][j-1][a[i]] ) dp[i+1][j][k][l]+=me;
            else {
                if ( k>0 ) dp[i+1][j][k-1][l]+=me*k;
                dp[i][j][k][l+1]+=me;
            }
        } else if ( l==1 ) {
            if ( has[i-1][j-1][b[j]] ) dp[i][j+1][k][l]+=me;
            else {
                if ( a[i]!=b[j] && k>0 ) dp[i][j+1][k-1][l]+=me*k;
                dp[i][j][k][l+1]+=me;
            }
        } else if ( l==2 ) {
            if ( a[i]!=b[j] ) dp[i+1][j+1][k+1][0]+=me;
        }
    }
    // dump(ans);
    REP1(i,1,n) if ( i%3!=0 ) ans*=i;
    W(ans);
}

// {{{ main
}}
int main() { shik::main(); return 0; }
// }}}