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;

using ll = long long;
using pii = pair<int, int>;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i, n) rep2(i, 0, n)
#define rep2(i, m, n) for (int i = m; i < (n); i++)
#define ALL(c) (c).begin(), (c).end()

constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }

template <class T, class U>
void chmin(T& t, const U& u) {
    if (t > u) t = u;
}
template <class T, class U>
void chmax(T& t, const U& u) {
    if (t < u) t = u;
}

template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
    os << "(" << p.first << "," << p.second << ")";
    return os;
}

template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "{";
    rep(i, v.size()) {
        if (i) os << ",";
        os << v[i];
    }
    os << "}";
    return os;
}

#ifdef LOCAL
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << H;
    debug_out(T...);
}
#define debug(...) \
    cerr << __LINE__ << " [" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif

template <unsigned int MOD>
struct ModInt {
    using uint = unsigned int;
    using ull = unsigned long long;
    using M = ModInt;

    uint v;

    ModInt(ll _v = 0) { set_norm(_v % MOD + MOD); }
    M& set_norm(uint _v) {  //[0, MOD * 2)->[0, MOD)
        v = (_v < MOD) ? _v : _v - MOD;
        return *this;
    }

    explicit operator bool() const { return v != 0; }
    M operator+(const M& a) const { return M().set_norm(v + a.v); }
    M operator-(const M& a) const { return M().set_norm(v + MOD - a.v); }
    M operator*(const M& a) const { return M().set_norm(ull(v) * a.v % MOD); }
    M operator/(const M& a) const { return *this * a.inv(); }
    M& operator+=(const M& a) { return *this = *this + a; }
    M& operator-=(const M& a) { return *this = *this - a; }
    M& operator*=(const M& a) { return *this = *this * a; }
    M& operator/=(const M& a) { return *this = *this / a; }
    M operator-() const { return M() - *this; }
    M& operator++(int) { return *this = *this + 1; }
    M& operator--(int) { return *this = *this - 1; }

    M pow(ll n) const {
        if (n < 0) return inv().pow(-n);
        M x = *this, res = 1;
        while (n) {
            if (n & 1) res *= x;
            x *= x;
            n >>= 1;
        }
        return res;
    }

    M inv() const {
        ll a = v, b = MOD, p = 1, q = 0, t;
        while (b != 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(p -= t * q, q);
        }
        return M(p);
    }

    bool operator==(const M& a) const { return v == a.v; }
    bool operator!=(const M& a) const { return v != a.v; }
    friend ostream& operator<<(ostream& os, const M& a) { return os << a.v; }
    static uint get_mod() { return MOD; }
};

using Mint = ModInt<1000000007>;
const int maxn = 410;

Mint dp[2][maxn][maxn];  // count mod 2, index of a and b
Mint ac[2][maxn][maxn];
int num[maxn][maxn];
bool valid[maxn][maxn];

int main() {
    int N;
    cin >> N;
    V<int> a(N), b(N);
    rep(i, N) cin >> a[i], --a[i];
    rep(i, N) cin >> b[i], --b[i];

    rep(i, N) {
        rep(j, N) {
            num[i][j] = 0;
            V<bool> used(N);
            rep(k, i + 1) { used[a[k]] = true; }
            rep(k, j + 1) { used[b[k]] = true; }
            int cnt = 0;
            rep(k, N) if (used[k]) { cnt++; }
            num[i][j] = cnt;
        }
    }

    rep(i, N) {
        rep(j, N) {
            bool f = a[i] != b[j];
            rep(k, j) if (b[k] == a[i]) f = false;
            rep(k, i) if (a[k] == b[j]) f = false;
            valid[i][j] = f;
        }
    }

    int n = N / 3;

    int c = 0, f = 1;

    dp[c][0][0] = 1;

    rep(i, n) {
        rep(j, maxn) {
            rep(k, maxn) { dp[f][j][k] = ac[f][j][k] = 0; }
        }

        rep(j, N) {
            rep(k, N) {
                ac[f][j + 1][k + 1] = ac[f][j + 1][k] + ac[f][j][k + 1] -
                                      ac[f][j][k] + dp[c][j][k];
                if (!valid[j][k]) continue;
                int free = N - num[j][k] - (n - i - 1) * 3;
                if (free > 0) {
                    dp[f][j + 1][k + 1] += ac[f][j + 1][k + 1] * free;
                }
            }
        }
        swap(c, f);
    }

    Mint ans = 0;
    rep(j, N + 1) {
        rep(k, N + 1) { ans += dp[c][j][k]; }
    }

    Mint coef(1);
    rep(i, n) {
        int num = i * 3;
        coef *= (num + 1) * (num + 2);
    }

    debug(ans, coef);
    cout << ans * coef << endl;

    return 0;
}