// Original C++ solution (slower):
#include <algorithm>

#include <bitset>

#include <complex>

#include <deque>

#include <iostream>

#include <istream>

#include <iterator>

#include <map>

#include <queue>

#include <set>

#include <stack>

#include <string>

#include <vector>

#include <tuple>

#include <iomanip>

using namespace std;



typedef long long ll;

typedef unsigned long long ull;

//typedef pair<ll, ll> P;

typedef vector<ll> vec;

typedef vector<vec> mat;

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

#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)

#define pb push_back

#define f first

#define s second

/*

ll max(ll a, ll b){return (a > b) ? a : b;}

ll min(ll a, ll b){return (a < b) ? a : b;}

ll max3(ll a, ll b, ll c){return max(a, max(b, c));};

ll min3(ll a, ll b, ll c){return min(a, min(b, c));};

ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};

ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};

ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};

ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};

*/

const ll INFL = 1LL << 60;//10^18 = 2^60

//const int INF = 1 << 30;//10^9

const ll INF = INFL;

//ll MOD = 1000000007;

ll MOD = 998244353;



vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};

vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};



//デバッグ

void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}



ll pow_long(ll x, ll k){

  ll res = 1;

  while(k > 0){

    if(k % 2) res *= x;

    x *= x;

    k /= 2;

  }

  return res;

}



ll pow_mod(ll x, ll k){

  ll res = 1;

  while(k > 0){

    if(k % 2){

      res *= x; res %= MOD;

    }

    x *= x; x %= MOD;

    k /= 2;

  }

  return res;

}



ll inverse(ll x){return pow_mod(x, MOD - 2);};



ll gcd(ll a, ll b){

    if(b == 0) return a;

    return gcd(b, a % b);

}



ll lcm(ll x, ll y){return x / gcd(x, y) * y;};



ll kai_mod(ll x){

  if(x == 0) return 1;

  return x * kai_mod(x-1) % MOD;

}



//約数の列挙O(√n)

vector<ll> divisor(ll n){

    vector<ll> res(0);

    for(ll i = 1; i * i <= n; i++){

        if(n % i == 0){

            res.push_back(i);

            if(i != n/i) res.push_back(n/i);

        }

    }

    sort(res.begin(), res.end());

    return res;

}



//素因数列挙

vector<ll> MakePrimeList(ll x) {

    vector<ll> res;

    for(ll k=2; k*k<=x; k++) {

        if(x % k == 0) {

            res.pb(k);

            while(x % k == 0) x /= k;

        }

    }

    if(x > 1) res.push_back(x);

    return res;

}





//コンビネーション

const int MAXcomb = 200010;

ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];

//facはn!,finvは1/n!

//invは逆元

void COMinit(){

    fac[0] = fac[1] = 1;

    finv[0] = finv[1] = 1;

    inv[1] = 1;

    for(int i = 2; i < MAXcomb; 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 comb(int n, int k){

    if(n < k) return 0;

    if(n < 0 || k < 0) return 0;

    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;

}



ll N;

long double x[100010], y[100010];

char d[100010];

long double uu = -INF, ud = -INF, ut = -INF;//上限,増加

long double du = INF, dd = INF, dt = INF;

long double lu = INF, ld = INF, lt = INF;

long double ru = -INF, rd = -INF, rt = -INF;

vector<double> point;



void add(long double d, long double t, long double u){

  if(abs(u) < INF/2 && abs(t) < INF/2){

    if(t-u >= 0) point.pb(t-u);

  }

  if(abs(u) < INF/2 && abs(d) < INF/2){

    if(d-u >= 0) point.pb((d-u)/2.0);

  }

  if(abs(d) < INF/2 && abs(t) < INF/2){

    if(d-t >= 0) point.pb(d-t);

  }

}



long double get(long double t){

  long double u = -INF, d = INF, l = INF, r = -INF;

  if(abs(uu) < INF/2) u = max(u, uu+t);

  if(abs(ut) < INF/2) u = max(u, ut);

  if(abs(ud) < INF/2) u = max(u, ud-t);

  if(abs(ru) < INF/2) r = max(r, ru+t);

  if(abs(rt) < INF/2) r = max(r, rt);

  if(abs(rd) < INF/2) r = max(r, rd-t);

  if(abs(du) < INF/2) d = min(d, du+t);

  if(abs(dt) < INF/2) d = min(d, dt);

  if(abs(dd) < INF/2) d = min(d, dd-t);

  if(abs(lu) < INF/2) l = min(l, lu+t);

  if(abs(lt) < INF/2) l = min(l, lt);

  if(abs(ld) < INF/2) l = min(l, ld-t);

  return (r-l) * (u-d);

}



long double solve(long double s, long double t){

  long double left = s, right = t;

  for(ll times = 0; times < 1000; times++){

    long double mid1 = left + (right - left) / 3;

    long double mid2 = left + (right - left) / 3 * 2;

    long double x1 = get(mid1);

    long double x2 = get(mid2);

    if(x1 < x2){

      right = mid2;

    }else{

      left = mid1;

    }

  }

  long double res = get(left);

  res = min(res, get(s));

  res = min(res, get(t));

  return res;

}



int main(){

  cin >> N;

  rep(i, N){

    cin >> x[i] >> y[i] >> d[i];

    if(d[i] == 'U'){

      uu = max(uu, y[i]);

      du = min(du, y[i]);

      lt = min(lt, x[i]);

      rt = max(rt, x[i]);

    }else if(d[i] == 'D'){

      ud = max(ud, y[i]);

      dd = min(dd, y[i]);

      lt = min(lt, x[i]);

      rt = max(rt, x[i]);

    }else if(d[i] == 'L'){

      ut = max(ut, y[i]);

      dt = min(dt, y[i]);

      ld = min(ld, x[i]);

      rd = max(rd, x[i]);

    }else{

      ut = max(ut, y[i]);

      dt = min(dt, y[i]);

      lu = min(lu, x[i]);

      ru = max(ru, x[i]);

    }

  }



  add(ud, ut, uu);

  add(dd, dt, du);

  add(ld, lt, lu);

  add(rd, rt, ru);

  point.pb(0);

  point.pb(1e12);

  sort(point.begin(), point.end());

  ll size = point.size();

  long double ans = INFL;

  rep(i, size-1){

    ans = min(ans, solve(point[i], point[i+1]));

  }

  cout << fixed << setprecision(20) << ans << endl;



}


// Optimized solution to the same problem (faster):
#include <algorithm>

#include <bitset>

#include <complex>

#include <deque>

#include <iostream>

#include <istream>

#include <iterator>

#include <map>

#include <queue>

#include <set>

#include <stack>

#include <string>

#include <vector>

#include <tuple>

#include <iomanip>

using namespace std;



typedef long long ll;

typedef unsigned long long ull;

//typedef pair<ll, ll> P;

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

#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)

#define pb push_back

#define f first

#define s second

#define chmin(x, y) x = min(x, y);

#define chmax(x, y) x = max(x, y);

/*

ll max(ll a, ll b){return (a > b) ? a : b;}

ll min(ll a, ll b){return (a < b) ? a : b;}

ll max3(ll a, ll b, ll c){return max(a, max(b, c));};

ll min3(ll a, ll b, ll c){return min(a, min(b, c));};

ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};

ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};

ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};

ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};

*/

const ll INFL = 1LL << 60;//10^18 = 2^60

//const int INF = 1 << 30;//10^9

const ll INF = INFL;

ll MOD = 1000000007;

//ll MOD = 998244353;



vector<ll> dy = {0, -1, 0, 1, 1, 1, -1, -1, 0};

vector<ll> dx = {1, 0, -1, 0, 1, -1, 1, -1, 0};



void pres(double A, ll x = 10){cout << fixed << setprecision(x) << A << endl;}

//デバッグ

void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}

ll get_bit(ll x){return __builtin_popcountll(x);}



ll pow_long(ll x, ll k){

  ll res = 1;

  while(k > 0){

    if(k % 2) res *= x;

    x *= x;

    k /= 2;

  }

  return res;

}



ll pow_mod(ll x, ll k){

  ll res = 1;

  while(k > 0){

    if(k % 2){

      res *= x; res %= MOD;

    }

    x *= x; x %= MOD;

    k /= 2;

  }

  return res;

}



ll inverse(ll x){return pow_mod(x, MOD - 2);};



ll gcd(ll a, ll b){

    if(b == 0) return a;

    return gcd(b, a % b);

}



ll lcm(ll x, ll y){return x / gcd(x, y) * y;};



ll kai_mod(ll x){

  if(x == 0) return 1;

  return x * kai_mod(x-1) % MOD;

}



//約数の列挙O(√n)

vector<ll> divisor(ll n){

    vector<ll> res(0);

    for(ll i = 1; i * i <= n; i++){

        if(n % i == 0){

            res.push_back(i);

            if(i != n/i) res.push_back(n/i);

        }

    }

    sort(res.begin(), res.end());

    return res;

}



//素因数列挙

vector<ll> MakePrimeList(ll x) {

    vector<ll> res;

    for(ll k=2; k*k<=x; k++) {

        if(x % k == 0) {

            res.pb(k);

            while(x % k == 0) x /= k;

        }

    }

    if(x > 1) res.push_back(x);

    return res;

}







//コンビネーション

const int MAXcomb = 2000100;

ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];

//facはn!,finvは1/n!

//invは逆元

void COMinit(){

    fac[0] = fac[1] = 1;

    finv[0] = finv[1] = 1;

    inv[1] = 1;

    for(int i = 2; i < MAXcomb; 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 comb(int n, int k){

    if(n < k) return 0;

    if(n < 0 || k < 0) return 0;

    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;

}

ll disit(ll x){

  ll res = 0;

  while(x){

    res++;

    x /= 10;

  }

  return res;

}



double get_t(double a1, double b1, double a2, double b2){

  return (b2 - b1) / (a1 - a2);

}



double LL = INFL, LR = -INFL, TL = INFL, TR = -INFL, RL = INFL, RR = -INFL;

double DD = INFL, DU = -INFL, TD = INFL, TU = -INFL, UD = INFL, UU = -INFL;



double X(double t){

  double mX = min(min(LL - t, RL + t), TL);

  double MX = max(max(LR - t, RR + t), TR);

  return MX - mX;

}



double Y(double t){

  double mY = min(min(DD - t, UD + t), TD);

  double MY = max(max(DU - t, UU + t), TU);

  return MY - mY;

}



double subsolve(double s, double t){

  //時間を求める

  double left = s, right = t;

  rep(t, 1000){

    double mid1 = (left * 2 + right) / 3;

    double mid2 = (left + right * 2) / 3;

    double x1 = X(mid1);

    double y1 = Y(mid1);

    double x2 = X(mid2);

    double y2 = Y(mid2);

    if(x1 * y1 >= x2 * y2){

      left = mid1;

    }else{

      right = mid2;

    }

  }

  if(left < 0) left = 0;

  return X(left) * Y(left);

}



void solve(){

  ll N;

  cin >> N;

  rep(i, N){

    double x, y;

    char d;

    cin >> x >> y >> d;

    if(d == 'D' || d == 'U'){

      chmin(TL, x);

      chmax(TR, x);

      if(d == 'D'){

        chmin(DD, y);

        chmax(DU, y);

      }else{

        chmin(UD, y);

        chmax(UU, y);

      }

    }else{

      chmin(TD, y);

      chmax(TU, y);

      if(d == 'L'){

        chmin(LL, x);

        chmax(LR, x);

      }else{

        chmin(RL, x);

        chmax(RR, x);

      }

    }

  }



  double ans = (double)INFL;

  vector<pair<double, double>> xy;

  xy.pb({-1, LL});

  xy.pb({-1, LR});

  xy.pb({-1, DD});

  xy.pb({-1, DU});

  xy.pb({0, TL});

  xy.pb({0, TR});

  xy.pb({0, TD});

  xy.pb({0, TU});

  xy.pb({1, RL});

  xy.pb({1, RR});

  xy.pb({1, UD});

  xy.pb({1, UU});

  vector<double> T;

  rep(i, xy.size()){

    rep(j, xy.size()){

      if(i >= j) continue;

      if(xy[i].f == xy[j].f) continue;

      if(abs(xy[i].s) >= INF || abs(xy[j].s) >= INF) continue;

      T.pb(get_t(xy[i].f, xy[i].s, xy[j].f, xy[j].s));

    }

  }



  sort(T.begin(), T.end());

  rep(i, T.size() - 1){

    chmin(ans, subsolve(T[i], T[i+1]));

  }

  pres(ans, 10);

}



int main(){

  cin.tie(0);

  ios::sync_with_stdio(false);

  solve();

}


// Fastest known solution for this problem: