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>
#define de(x) cout<<#x<<"="<<x<<endl
#define dd(x) cout<<#x<<"="<<x<<" "
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define all(a) a.begin(),a.end()
using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> pll;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
typedef pair<int,pii> piii;

const ll mod = 1000000007;
const int N = 2e5+6;
const int M = 2e6+6;
const double eps = 1e-12;
const ll inf = 1e16;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int n;
db X[N], Y[N];
int D[N];
db cal(db t) {
	db mxx = -inf, mnx = inf, mxy = -inf, mny = inf;
	rep(i, 0, n) {
		db nx = X[i], ny = Y[i];
		if(D[i]==1) ny += t;
		if(D[i]==3) ny -= t;
		if(D[i]==2) nx += t;
		if(D[i]==4) nx -= t;
		mxx = max(nx, mxx);
		mnx = min(nx, mnx);
		mxy = max(ny, mxy);
		mny = min(ny, mny);
	}
	return (mxx - mnx) * (mxy - mny);
}
db three_search(db l, db r) {
	double lm, rm;
	int cnt = 0;
    while (r > eps + l && cnt <= 500) {
    	// dd(left);de(right);
        lm = (l + r) / 2;
        rm = (lm + r) / 2;
        // 如果是求最小值的话这里判<即可
        if(cal(lm) < cal(rm)) r = rm;
        else l = lm;
        cnt++;
    }
    return cal(l);
}
int main() {
	// dp[i][j] 表示以s[i]和t[j]结尾的相同子序列的个数
	// 到后面就是前缀和了
	while(~scanf("%d", &n)) {
		rep(i, 0, n) {
			char c;
			scanf("%lf%lf %c", X + i, Y + i, &c);
			if(c=='U') D[i] = 1;
			if(c=='R') D[i] = 2;
			if(c=='D') D[i] = 3;
			if(c=='L') D[i] = 4;
		}
		printf("%.10f\n", three_search(0, 1e12));
	}
    return 0;
}