#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define int long long
#define st first
#define nd second
#define rd third
#define FOR(i, a, b) for(int i =(a); i <=(b); ++i)
#define RE(i, n) FOR(i, 1, n)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define REP(i, n) for(int i = 0;i <(n); ++i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
using namespace std;
template<typename TH> void _dbg(const char* sdbg, TH h) { cerr<<sdbg<<"="<<h<<"\n"; }
template<typename TH, typename... TA> void _dbg(const char* sdbg, TH h, TA... t) {
  while(*sdbg != ',')cerr<<*sdbg++; cerr<<"="<<h<<","; _dbg(sdbg+1, t...);
}
#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#define debugv(x) {{cerr <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<"\n"; }}
#else
#define debug(...) (__VA_ARGS__)
#define debugv(x)
#define cerr if(0)cout
#endif
#define next ____next
#define prev ____prev
#define left ____left
#define hash ____hash
typedef long long ll;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<pair<int, int> > VPII;
typedef vector<pair<ll, ll> > VPLL;

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
template<class T1, class T2>
ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class A, class B, class C> struct Triple { A first; B second; C third;
  bool operator<(const Triple& t) const { if (st != t.st) return st < t.st; if (nd != t.nd) return nd < t.nd; return rd < t.rd; } };
template<class T> void ResizeVec(T&, vector<int>) {}
template<class T> void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]); sz.erase(sz.begin()); if (sz.empty()) { return; }
  for (T& v : vec) { ResizeVec(v, sz); }
}
typedef Triple<int, int, int> TIII;
template<class A, class B, class C>
ostream& operator<< (ostream &out, Triple<A, B, C> t) { return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")"; }
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }


// dp[i][j][k] - A zjadł ity, B zjadl jty, bylo k tur
// kazdy ziomek ktory byl przed Ai lub Bj zostal juz zjedzony
// kazdy z A, B, C zje jeszcze (n/3-k)
// jak przed Ai oraz Bj jest x ziomkow to znaczy, ze C zjadl
// dowolne x - 3(n/3-k)

const int P = 1e9 + 7;

int spow(int a, int b) {
  int r = 1;
  while (b) {
    if (b % 2) {
      r = r * a % P;
    }
    a = a * a % P;
    b /= 2;
  }
  return r;
}
int Inv(int a) {
  return spow(a, P - 2);
}
const int N = 405;
int inv[N];
int a[N];
int b[N];
int dp[N][N][2][2];
int past[N][N]; // oba dalej niz i oraz j
int wh[N][2];
int fact[N];

int32_t main() {

  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  cin.tie(0);
  //double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  
  int n;
  cin>>n;
  fact[0] = 1;
  RE (i, n) {
    inv[i] = Inv(i);
    fact[i] = i * fact[i - 1] % P;
  }
  RE (i, n) {
    cin>>a[i];
    wh[a[i]][0] = i;
  }
  RE (i, n) {
    cin>>b[i];
    wh[b[i]][1] = i;
  }
  RE (i, n) {
    RE (j, n) {
      if (a[i] == b[j]) {
        past[i - 1][j - 1] = 1;
      }
    }
  }
  FORD (i, n, 0) {
    FORD (j, n, 0) {
      past[i][j] = past[i + 1][j] + past[i][j + 1] + past[i][j] - past[i + 1][j + 1];
    }
  }
  if (a[1] == b[1]) {
    cout<<"0\n";
    return 0;
  }
  assert(past[1][1] == n - 2);
  dp[1][1][0][0] = n - 2;
  int nxt = 1;
  //dp[i][j][who][prv] ->
  //  k == 0
  //    -> (i+1
  FOR (k, 2, n / 3) {
    RE (i, n) {
      RE (j, n) {
        REP (who, 2) {
          dp[i][j][who][nxt] = 0;
        }
      }
    }
    int cur = nxt ^ 1;
    RE (i, n) {
      RE (j, n) {
        cerr<<dp[i][j][0][cur]<<" ";
      }
      cerr<<endl;
    }
    cerr<<endl;
    RE (i, n) {
      RE (j, n) {
        int x = past[i][j];
        // x to 3(n/3-(k-1)) + |tyle ile zjadl juz C| w sumie do zjedzenia przez A, B, C
        int Cate = x - 3 * (n / 3 - (k - 1));
        if (Cate < 0) { continue; }
        int Y = dp[i][j][0][cur];
        if (Y) {
          debug(i, j, Y, wh[a[i]][1] > j);
        }
        if (wh[a[i + 1]][1] > j) { // nie bylo
          //if (k
          int& D = dp[i + 1][j][0][cur];
          D = (D + Y * Cate % P * inv[x]) % P;
          int& E = dp[i + 1][j][1][cur];
          E = (E + Y * (x - Cate) % P * inv[x] % P) % P;
        } else {
          int& D = dp[i + 1][j][0][cur];
          D = (D + Y) % P;
        }
      }
    }
    RE (i, n) {
      RE (j, n) {
        cerr<<dp[i][j][1][cur]<<" ";
      }
      cerr<<endl;
    }
    RE (i, n) {
      RE (j, n) {
        int x = past[i][j];
        int Cate = x - (n / 3 - k) - (n / 3 - (k - 1)) - (n / 3 - (k - 1));
        if (Cate < 0) { continue; }
        int& Y = dp[i][j][1][cur];
        if (wh[b[j + 1]][0] > i) {
          int& D = dp[i][j + 1][1][cur];
          D = (D + Y * Cate % P * inv[x]) % P;
          int& E = dp[i][j + 1][0][nxt];
          E = (E + Y * (x - Cate) % P * inv[x] % P * (x - 1 - Cate)) % P;
        } else {
          if (wh[b[j + 1]][0] < i) {
            int& D = dp[i][j + 1][1][cur];
            D = (D + Y) % P;
          }
        }
      }
    }
    nxt ^= 1;
  }
  
  
  
  cerr<<endl;
  int res = 0;
  RE (i, n) {
    RE (j, n) {
      //int x = past[i][j];
      cerr<<dp[i][j][0][nxt ^ 1]<<" ";
      res = (res + dp[i][j][0][nxt ^ 1] /** fact[x]*/) % P;
    }
    cerr<<endl;
  }
  debug(res);
  
  
  
  RE (i, n / 3) {
    res = 1ll * res * (3 * i - 2) % P;
    res = 1ll * res * (3 * i - 1) % P;
  }
  cout<<res<<endl;
  return 0;
}

### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define int long long
#define st first
#define nd second
#define rd third
#define FOR(i, a, b) for(int i =(a); i <=(b); ++i)
#define RE(i, n) FOR(i, 1, n)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define REP(i, n) for(int i = 0;i <(n); ++i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
using namespace std;
template<typename TH> void _dbg(const char* sdbg, TH h) { cerr<<sdbg<<"="<<h<<"\n"; }
template<typename TH, typename... TA> void _dbg(const char* sdbg, TH h, TA... t) {
  while(*sdbg != ',')cerr<<*sdbg++; cerr<<"="<<h<<","; _dbg(sdbg+1, t...);
}
#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#define debugv(x) {{cerr <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<"\n"; }}
#else
#define debug(...) (__VA_ARGS__)
#define debugv(x)
#define cerr if(0)cout
#endif
#define next ____next
#define prev ____prev
#define left ____left
#define hash ____hash
typedef long long ll;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<pair<int, int> > VPII;
typedef vector<pair<ll, ll> > VPLL;

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
template<class T1, class T2>
ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class A, class B, class C> struct Triple { A first; B second; C third;
  bool operator<(const Triple& t) const { if (st != t.st) return st < t.st; if (nd != t.nd) return nd < t.nd; return rd < t.rd; } };
template<class T> void ResizeVec(T&, vector<int>) {}
template<class T> void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]); sz.erase(sz.begin()); if (sz.empty()) { return; }
  for (T& v : vec) { ResizeVec(v, sz); }
}
typedef Triple<int, int, int> TIII;
template<class A, class B, class C>
ostream& operator<< (ostream &out, Triple<A, B, C> t) { return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")"; }
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }


// dp[i][j][k] - A zjadł ity, B zjadl jty, bylo k tur
// kazdy ziomek ktory byl przed Ai lub Bj zostal juz zjedzony
// kazdy z A, B, C zje jeszcze (n/3-k)
// jak przed Ai oraz Bj jest x ziomkow to znaczy, ze C zjadl
// dowolne x - 3(n/3-k)

const int P = 1e9 + 7;

int spow(int a, int b) {
  int r = 1;
  while (b) {
    if (b % 2) {
      r = r * a % P;
    }
    a = a * a % P;
    b /= 2;
  }
  return r;
}
int Inv(int a) {
  return spow(a, P - 2);
}
const int N = 405;
int inv[N];
int a[N];
int b[N];
int dp[N][N][2][2];
int past[N][N]; // oba dalej niz i oraz j
int wh[N][2];
int fact[N];

int32_t main() {

  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  cin.tie(0);
  //double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  
  int n;
  cin>>n;
  fact[0] = 1;
  RE (i, n) {
    inv[i] = Inv(i);
    fact[i] = i * fact[i - 1] % P;
  }
  RE (i, n) {
    cin>>a[i];
    wh[a[i]][0] = i;
  }
  RE (i, n) {
    cin>>b[i];
    wh[b[i]][1] = i;
  }
  RE (i, n) {
    RE (j, n) {
      if (a[i] == b[j]) {
        past[i - 1][j - 1] = 1;
      }
    }
  }
  FORD (i, n, 0) {
    FORD (j, n, 0) {
      past[i][j] = past[i + 1][j] + past[i][j + 1] + past[i][j] - past[i + 1][j + 1];
    }
  }
  if (a[1] == b[1]) {
    cout<<"0\n";
    return 0;
  }
  assert(past[1][1] == n - 2);
  dp[1][1][0][0] = n - 2;
  int nxt = 1;
  //dp[i][j][who][prv] ->
  //  k == 0
  //    -> (i+1
  FOR (k, 2, n / 3) {
    RE (i, n) {
      RE (j, n) {
        REP (who, 2) {
          dp[i][j][who][nxt] = 0;
        }
      }
    }
    int cur = nxt ^ 1;
    RE (i, n) {
      RE (j, n) {
        cerr<<dp[i][j][0][cur]<<" ";
      }
      cerr<<endl;
    }
    cerr<<endl;
    RE (i, n) {
      RE (j, n) {
        int x = past[i][j];
        // x to 3(n/3-(k-1)) + |tyle ile zjadl juz C| w sumie do zjedzenia przez A, B, C
        int Cate = x - 3 * (n / 3 - (k - 1));
        if (Cate < 0) { continue; }
        int Y = dp[i][j][0][cur];
        if (Y) {
          debug(i, j, Y, wh[a[i]][1] > j);
        }
        if (wh[a[i + 1]][1] > j) { // nie bylo
          //if (k
          int& D = dp[i + 1][j][0][cur];
          D = (D + Y * Cate % P * inv[x]) % P;
          int& E = dp[i + 1][j][1][cur];
          E = (E + Y * (x - Cate) % P * inv[x] % P) % P;
        } else {
          int& D = dp[i + 1][j][0][cur];
          D = (D + Y) % P;
        }
      }
    }
    RE (i, n) {
      RE (j, n) {
        cerr<<dp[i][j][1][cur]<<" ";
      }
      cerr<<endl;
    }
    RE (i, n) {
      RE (j, n) {
        int x = past[i][j];
        int Cate = x - (n / 3 - k) - (n / 3 - (k - 1)) - (n / 3 - (k - 1));
        if (Cate < 0) { continue; }
        int& Y = dp[i][j][1][cur];
        if (wh[b[j + 1]][0] > i) {
          int& D = dp[i][j + 1][1][cur];
          D = (D + Y * Cate % P * inv[x]) % P;
          int& E = dp[i][j + 1][0][nxt];
          E = (E + Y * (x - Cate) % P * inv[x] % P * (x - 1 - Cate)) % P;
        } else {
          if (wh[b[j + 1]][0] < i) {
            int& D = dp[i][j + 1][1][cur];
            D = (D + Y) % P;
          }
        }
      }
    }
    nxt ^= 1;
  }
  
  
  
  cerr<<endl;
  int res = 0;
  RE (i, n) {
    RE (j, n) {
      //int x = past[i][j];
      cerr<<dp[i][j][0][nxt ^ 1]<<" ";
      res = (res + dp[i][j][0][nxt ^ 1] /** fact[x]*/) % P;
    }
    cerr<<endl;
  }
  debug(res);
  
  
  
  RE (i, n / 3) {
    res = 1ll * res * (3 * i - 2) % P;
    res = 1ll * res * (3 * i - 1) % P;
  }
  cout<<res<<endl;
  return 0;
}

```