Problem:
Score : 500 points

Problem Statement

There are 2N squares arranged from left to right. You are given a string of length 2N representing the color of each of the squares.

The color of the i-th square from the left is black if the i-th character of S is B, and white if that character is W.

You will perform the following operation exactly N times: choose two distinct squares, then invert the colors of these squares and the squares between them. Here, to invert the color of a square is to make it white if it is black, and vice versa.

Throughout this process, you cannot choose the same square twice or more. That is, each square has to be chosen exactly once.

Find the number of ways to make all the squares white at the end of the process, modulo 10^9+7.

Two ways to make the squares white are considered different if and only if there exists i (1 \leq i \leq N) such that the pair of the squares chosen in the i-th operation is different.

Constraints

1 \leq N \leq 10^5

|S| = 2N

Each character of S is B or W.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of ways to make all the squares white at the end of the process, modulo 10^9+7. If there are no such ways, print 0.

Sample Input 1

2
BWWB

Sample Output 1

4

There are four ways to make all the squares white, as follows:

Choose Squares 1, 3 in the first operation, and choose Squares 2, 4 in the second operation.

Choose Squares 2, 4 in the first operation, and choose Squares 1, 3 in the second operation.

Choose Squares 1, 4 in the first operation, and choose Squares 2, 3 in the second operation.

Choose Squares 2, 3 in the first operation, and choose Squares 1, 4 in the second operation.

Sample Input 2

4
BWBBWWWB

Sample Output 2

288

Sample Input 3

5
WWWWWWWWWW

Sample Output 3

0

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;
using lb = long double;
using P = pair<ll,ll>;
using graph = vector<vector<ll>>;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ll INF = 1LL<<60;
const ll mod = 1000000007LL;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)

struct mint {
    long long x;
public:
    mint(long long x=0) : x((x%mod+mod)%mod) {}
    mint operator-() const { 
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    bool operator <(const mint& r){return this->x<r.x;}
    bool operator ==(const mint& r){return this->x==r.x;}
    bool operator !=(const mint& r){return this->x!=r.x;}
    void operator++() {
        x++;
    }
    void operator--() {
        x--;
        if(x<0) x += mod;
    }
    // for prime mod
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }

    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
};

int main(){
  mint N;
  cin>>N.x;
  string S;
  cin>>S;
  vector<int> v(N.x*2);
  rep(i,N.x*2){
    if(S[i] == 'B') v[i] = 0;
    else v[i] = 1;
  }
  mint ans = 0;
  if(v[0] == 0&& v[N.x*2-1] == 0){
    vector<int> t(N.x * 2);
    t[0] == 0;
    rep(i,N.x*2-1){
      if(v[i+1]!=v[i]) t[i+1] = t[i];
      else t[i+1] = 1-t[i];
    }
    ll check = 0;
    rep(i,N.x*2) check += t[i];
    if(check == N.x){
      if(t[N.x*2-1]==1) {
        mint cnt = 0;
        ans = 1;
        rep(i,N.x * 2){
          cnt += t[N.x*2-1-i];
          if(t[N.x*2-1-i]==0) {
            ans*=cnt;
            --cnt;
          }
        }
        rep(i,N.x) ans *= (mint)(i+1);
      }
    }
  }
  cout<<ans<<endl;
}