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:

//#define USEPB_DS
#define USETR1
#define CPPELEVEN
#define GPP

/*
 * temp.cpp
 *
 *  Created on: 2012-7-18
 *      Author: BSBandme
 */
//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <iostream>
#include <fstream>
#include <string.h>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <cassert>
#include <list>
#include <iomanip>
#include <math.h>
#include <deque>
#include <utility>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <climits>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <functional>
#include <sstream>
#include <tuple>

using namespace std;

#ifndef CPPELEVEN
#ifdef USETR1
#include <tr1/unordered_map>
#include <tr1/unordered_set>
using namespace tr1;
#endif
#else
#include <unordered_map>
#include <unordered_set>
#endif

#ifdef USEPB_DS
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
// binomial_heap_tag, rc_binomial_heap_tag, thin_heap_tag, binary_heap_tag
typedef __gnu_pbds::priority_queue<int, greater<int>, pairing_heap_tag> pq_type;
// splay_tree_tag, ov_tree_tag
typedef tree <int, null_type, less <int>, rb_tree_tag, tree_order_statistics_node_update> tree_type;
#endif

#define mpr make_pair
typedef unsigned int ui;
typedef unsigned long long ull;
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <double, double> pdd;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <double> vd;
typedef vector <string> vs;
typedef map <string, int> mpsi;
typedef map <double, int> mpdi;
typedef map <int, int> mpii;

const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};

template <class T> inline T abs1(T a) {return a < 0 ? -a : a;}

#ifndef CPPELEVEN
template <class T> inline T max1(T a, T b) { return b < a ? a : b; }
template <class T> inline T max1(T a, T b, T c) { return max1(max1(a, b), c); }
template <class T> inline T max1(T a, T b, T c, T d) { return max1(max1(a, b, c), d); }
template <class T> inline T max1(T a, T b, T c, T d, T e) { return max1(max1(a, b, c, d), e); }
template <class T> inline T min1(T a, T b) { return a < b ? a : b; }
template <class T> inline T min1(T a, T b, T c) { return min1(min1(a, b), c); }
template <class T> inline T min1(T a, T b, T c, T d) { return min1(min1(a, b, c), d); }
template <class T> inline T min1(T a, T b, T c, T d, T e) { return min1(min1(a, b, c, d), e); }
#else
template <typename t, typename t1>
t min1(t a, t1 b) { return a < b ? a : b; }
template <typename t, typename... arg>
t min1(t a, arg... arr) { return min1(a, min1(arr...)); }
template <typename t, typename t1>
t max1(t a, t1 b) { return a > b ? a : b; }
template <typename t, typename... arg>
t max1(t a, arg... arr) { return max1(a, max1(arr...)); }
#endif

inline int jud(double a, double b){
	if(abs(a) < eps && abs(b) < eps) return 0;
	else if(abs1(a - b) / max(abs1(a), abs1(b)) < eps) return 0;
	if(a < b) return -1;
	return 1;
}
template <typename t> inline int jud(t a, t b){
	if(a < b) return -1;
	if(a == b) return 0;
	return 1;
}

// f_lb == 1代表返回相同的一串的左边界，f_small == 1代表返回如果没有寻找的值返回小的数
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1){
	if(na == 0) return 0;
	int be = 0, en = na - 1;
	if(*a <= *(a + na - 1)){
		if(f_lb == 0) while(be < en){
			int mid = (be + en + 1) / 2;
			if(jud(*(a + mid), val) != 1) be = mid;
			else en = mid - 1;
		}else while(be < en){
			int mid = (be + en) / 2;
			if(jud(*(a + mid), val) != -1) en = mid;
			else be = mid + 1;
		}
		if(f_small && jud(*(a + be), val) == 1) be--;
		if(!f_small && jud(*(a + be), val) == -1) be++;
	} else {
		if(f_lb) while(be < en){
			int mid = (be + en + 1) / 2;
			if(jud(*(a + mid), val) != -1) be = mid;
			else en = mid - 1;
		}else while(be < en){
			int mid = (be + en) / 2;
			if(jud(*(a + mid), val) != 1) en = mid;
			else be = mid + 1;
		}
		if(!f_small && jud(*(a + be), val) == -1) be--;
		if(f_small && jud(*(a + be), val) == 1) be++;
	}
	return be;
}

template <class T> inline T lowb(T num) {return num & (-num); }
#ifdef GPP
inline int bitnum(ui nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(ull nValue) { return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32); }
inline int bitnum(ll nValue) { return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32); }
inline int bitmaxl(ui a) { if(a == 0) return 0; return 32 - __builtin_clz(a); }
inline int bitmaxl(int a) { if(a == 0) return 0; return 32 - __builtin_clz(a); }
inline int bitmaxl(ull a) { int temp = a >> 32; if(temp) return 32 - __builtin_clz(temp) + 32; return bitmaxl(int(a)); }
inline int bitmaxl(ll a) { int temp = a >> 32; if(temp) return 32 - __builtin_clz(temp) + 32; return bitmaxl(int(a)); }
#else
#endif

long long pow(long long n, long long m, long long mod = 0){
	if(m < 0) return 0;
	long long ans = 1;
	long long k = n;
	while(m){
		if(m & 1) {
			ans *= k;
			if(mod) ans %= mod;
		}
		k *= k;
		if(mod) k %= mod;
		m >>= 1;
	}
	return ans;
}

#define  MOD 1000000007
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
	if(mod == -1) mod = MOD;
	a += b;
	while(a >= mod) a -= mod;
	while(a < 0) a += mod;
}
template <class t>
void output1(t arr) {
	for(int i = 0; i < (int)arr.size(); i++)
		cerr << arr[i] << ' ';
	cerr << endl;
}
template <class t>
void output2(t arr) {
	for(int i = 0; i < (int)arr.size(); i++)
		output1(arr[i]);
}

//....................密..........封..........线..........下..........禁..........止..........hack...............................................

const int mod = MOD;
const int maxn = 410;
int uniq[maxn][maxn];
int have[maxn];
int n;
ll dp[maxn / 3][maxn][maxn];
int orig[2][maxn], rk[2][maxn];
ll fac[maxn];

int main() {


//............................不要再忘了检查maxn大小了！！！！BSBandme你个SB！！！！...................................................

	ios_base::sync_with_stdio(0);
	#ifdef DEBUG //......................................................................................................
	freopen("input.txt", "r", stdin);
	int __size__ = 256 << 20; // 256MB
	char *__p__ = (char*)malloc(__size__) + __size__;
	__asm__("movl %0, %%esp\n" :: "r"(__p__));
	#endif //...........................................................................................................

	fac[0] = 1;
	for (int i = 1; i < maxn; i++) {
		fac[i] = fac[i - 1] * i % mod;
	}
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", orig[0] + i);
		rk[0][orig[0][i]] = i;
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", orig[1] + i);
		rk[1][orig[1][i]] = i;
	}

	if (orig[1][0] == orig[0][0]) {
		cout << 0 << endl;
		return 0;
	}
	ll last = 1;
	for (int i = 0; i < n / 3; i++) {
		last = last * (i * 3 + 1) % mod * (i * 3 + 2) % mod;
	}

	uniq[0][0] = 0;
	for (int i = 0; i <= n; i++) {
		memset(have, 0, sizeof(have));
		uniq[i][0] = i;
		for (int j = 0; j < i; j++) {
			have[orig[0][j]] = 1;
		}
		for (int j = 0; j < n; j++) {
			uniq[i][j + 1] = uniq[i][j];
			if (!have[orig[1][j]]) {
				have[orig[1][j]] = 1;
				uniq[i][j + 1]++;
			}
		}
	}

	dp[0][0][0] = 1;
	for (int step = 0; step <= n / 3; step++) {
		for (int b = step; b <= step * 3; b++)
			for (int a = step;  uniq[a][b] <= step * 3 && a <= step * 3; a++) if (dp[step][a][b]) {
				if (rk[1][orig[0][a]] < b) {
					add(dp[step][a + 1][b], dp[step][a][b]);
				} else {
					ll can = step * 3 - uniq[a][b];
					if (can <= 0) {
						break;
					}
					add(dp[step][a + 1][b], 1ll * can * dp[step][a][b] % mod);
				}
			}
		for (int a = step; a <= step * 3; a++)
			for (int b = step;  uniq[a][b] <= step * 3 && b <= step * 3; b++) if (dp[step][a][b]) {
				if (rk[0][orig[1][b]] < a) {
					add(dp[step][a][b + 1], dp[step][a][b]);
				} else {
					ll can = step * 3 - uniq[a][b];
					if (can <= 0) {
						break;
					}
					add(dp[step][a][b + 1], 1ll * can * dp[step][a][b] % mod);
				}
			}
		if (step != n / 3) {
			for (int a = step; a <= step * 3; a++)
				for (int b = step; uniq[a][b] <= step * 3; b++)
					if (orig[0][a] != orig[1][b] && rk[0][orig[1][b]] >= a && rk[1][orig[0][a]] >= b)  {
						add(dp[step + 1][a + 1][b + 1], dp[step][a][b]);
					}
		}
	}
	cout << dp[n / 3][n][n] * last % mod << endl;

    return 0;
}