Problem:
Score : 600 points

Problem Statement

There are N points in a two-dimensional plane. The initial coordinates of the i-th point are (x_i, y_i). Now, each point starts moving at a speed of 1 per second, in a direction parallel to the x- or y- axis. You are given a character d_i that represents the specific direction in which the i-th point moves, as follows:

If d_i = R, the i-th point moves in the positive x direction;

If d_i = L, the i-th point moves in the negative x direction;

If d_i = U, the i-th point moves in the positive y direction;

If d_i = D, the i-th point moves in the negative y direction.

You can stop all the points at some moment of your choice after they start moving (including the moment they start moving).
Then, let x_{max} and x_{min} be the maximum and minimum among the x-coordinates of the N points, respectively. Similarly, let y_{max} and y_{min} be the maximum and minimum among the y-coordinates of the N points, respectively.

Find the minimum possible value of (x_{max} - x_{min}) \times (y_{max} - y_{min}) and print it.

Constraints

1 \leq N \leq 10^5

-10^8 \leq x_i,\ y_i \leq 10^8

x_i and y_i are integers.

d_i is R, L, U, or D.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1 d_1
x_2 y_2 d_2
.
.
.
x_N y_N d_N

Output

Print the minimum possible value of (x_{max} - x_{min}) \times (y_{max} - y_{min}).

The output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-9}.

Sample Input 1

2
0 3 D
3 0 L

Sample Output 1

0

After three seconds, the two points will meet at the origin. The value in question will be 0 at that moment.

Sample Input 2

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R

Sample Output 2

97.5

The answer may not be an integer.

Sample Input 3

20
6 -10 R
-4 -9 U
9 6 D
-3 -2 R
0 7 D
4 5 D
10 -10 U
-1 -8 U
10 -6 D
8 -5 U
6 4 D
0 3 D
7 9 R
9 -4 R
3 10 D
1 9 U
1 -6 U
9 -8 R
6 7 D
7 -3 D

Sample Output 3

273

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
//using namespace std;
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define rep(i,j,n) for(ll i=(ll)(j);i<(ll)(n);i++)
#define REP(i,j,n) for(ll i=(ll)(j);i<=(ll)(n);i++)
#define per(i,j,n) for(ll i=(ll)(j);(ll)(n)<=i;i--)
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(ll)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(ll)(key)))
#define pb emplace_back
#define mp std::make_pair
#define endl "\n"
#define double long double
using std::cin;
using std::cout;
using std::vector;
using std::string;
using std::upper_bound;
using std::lower_bound;
using vi=vector<ll>;
using vii=vector<vi>;
using pii=std::pair<ll,ll>;
constexpr ll MOD=1e9+7;
//constexpr ll MOD=998244353;
//constexpr ll MOD=100000;
constexpr ll MAX=1e7;
constexpr ll INF=(1ll<<60);
template<class T>
class prique :public std::priority_queue<T, std::vector<T>, std::greater<T>> {};
struct Binary_indexed_tree{
    int N;
    vi bit;
    Binary_indexed_tree(int n):N(n){
        bit.resize(N+1,0);
    }
    void add(int x,int a){
        for(x;x<=N;x+=(x&-x)) bit[x]+=a;
    }
    ll sum(int x){
        ll ret=0;
        for(x;x>0;x-=(x&-x)) ret+=bit[x];
        return ret;
    }
    ll lower_bound(ll X){
        if(sum(N)<X) return -1;
        ll ret=0,memo=1,sum=0;
        while(memo*2<=N) memo*=2;
        while(memo>0){
            if(memo+ret<=N&&sum+bit[memo+ret]<X){
                sum+=bit[memo+ret];
                ret+=memo;
            }
            memo/=2;
        }
        return ret+1;
    }
};
struct Union_Find{
    ll N;
    vi par;
    vi siz;
    Union_Find(int n):N(n){
        par.resize(N);
        siz.resize(N,1);
        rep(i,0,N) par[i]=i;
    }
    ll root(ll X){
        if(par[X]==X) return X;
        return par[X]=root(par[X]);
    }
    bool same(ll X,ll Y){
        return root(X)==root(Y);
    }
    void unite(ll X,ll Y){
        X=root(X);
        Y=root(Y);
        if(X==Y) return;
        par[X]=Y;
        siz[Y]+=siz[X];
        siz[X]=0;
    }
    ll size(ll X){
        return siz[root(X)];
    }
};
ll modpow(ll X,ll Y){
    ll sum=X,cnt=1;
    vi A;
    while(cnt<=Y){
        A.pb(sum);
        sum*=sum;
        sum%=MOD;
        cnt*=2;
    }
    int M=A.size();
    ll ret=1;
    REP(i,1,M){
        if(Y>=(1ll<<M-i)){
            Y-=(1ll<<M-i);
            ret*=A[M-i];
            ret%=MOD;
        }
    }
    return ret;
}
ll fac[MAX],finv[MAX],inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
ll COM(ll n,ll r){
    if(n<r||n<0||r<0) return 0;
    return fac[n]*finv[r]%MOD*finv[n-r]%MOD;
}
double len(double t,vi &A,vi &M){
    double min=MOD*3,max=0;
    rep(i,0,A.size()){
        double X=A[i]+M[i]*t;
        min=std::min(min,X);
        max=std::max(max,X);
    }
    return max-min;
}
signed main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    ll N; cin>>N;
    vii data(2,vi(N));
    vii memo(2,vi(N));
    rep(i,0,N){
        rep(j,0,2) cin>>data[j][i];
        char C; cin>>C;
        if(C=='L') memo[0][i]=-1;
        else if(C=='R') memo[0][i]=1;
        else if(C=='U') memo[1][i]=1;
        else memo[1][i]=-1;
    }
    vector<double> ans(2);
    rep(i,0,2){
        double left=0,right=1e10;
        rep(_,0,1000){
            double X=(right-left)/3.0;
            double Y=left+X,Z=left+X*2.0;
            if(len(Y,data[i],memo[i])<len(Z,data[i],memo[i])) right=Z;
            else left=Y;
        }
        ans[i]=len(left,data[0],memo[0])*len(left,data[1],memo[1]);
    }
    double ans_=std::min(ans[0],ans[1]);
    printf("%.15llf\n",ans_);
}