Problem: 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}.

Examples

Input

2
0 3 D
3 0 L


Output

0


Input

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


Output

97.5


Input

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


Output

273

Solution (in CPP):

Solution:
#pragma GCC optimize "O3"
#pragma GCC optimize "unroll-loops"
#pragma GCC target "sse,sse2,sse3,sse4,avx,abm,mmx,popcnt,tune=native"
#include<algorithm>
#include<cmath>
#include<cstdio>
typedef long double llf;
const int MN=1e5+4;
const llf INF=1e100,EPS=1e-12;
int N;
llf ans=INF;
struct pnt {
    llf x, y;
    llf ox,oy;
    char c;
}pnts[MN];
static inline void mov(pnt &p,llf dt) {
    switch(p.c) {
        case 'R':
            p.x=p.ox+dt;
            p.y=p.oy;
            break;
        case 'L':
            p.x=p.ox-dt;
            p.y=p.oy;
            break;
        case 'U':
            p.x=p.ox;
            p.y=p.oy+dt;
            break;
        case 'D':
            p.x=p.ox;
            p.y=p.oy-dt;
            break;
    }
}
static inline llf bb() {
    llf mxx=-INF,mxy=-INF,mnx=INF,mny=INF;
    for(int i=0;i<N;++i) {
        mxx=std::max(mxx,pnts[i].x);
        mnx=std::min(mnx,pnts[i].x);
        mxy=std::max(mxy,pnts[i].y);
        mny=std::min(mny,pnts[i].y);
    }
    return (mxx-mnx)*(mxy-mny);
}
int main() {
    scanf("%d",&N);
    for(int i=0;i<N;++i) {
        scanf("%Lf %Lf %c",&pnts[i].ox,&pnts[i].oy,&pnts[i].c);
    }
    // do ternary search
    llf lo=0,hi=1e9,mi1,mi2;
    for(int i=0;i<500;++i) {
        mi1=(2*lo+hi)/3;
        mi2=(lo+2*hi)/3;
        llf b1,b2;
        for(int i=0;i<N;++i)
            mov(pnts[i],mi1);
        b1=bb();
        for(int i=0;i<N;++i)
            mov(pnts[i],mi2);
        b2=bb();
        if(b1<b2) hi=mi2;
        else lo=mi1;
        ans=std::min(std::min(b1,b2),ans);
    }
    printf("%.9Lf\n",ans);
}