#include <bits/stdc++.h>
#define N 100005
using namespace std;

template <typename T> inline void cmax(T& x, T y) {
    if (x < y) x = y;
}

template <typename T> inline void cmin(T& x, T y) {
    if (x > y) x = y;
}

const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
int n, x[N], y[N], d[N];

double get(double now) {
    double xmin, xmax, ymin, ymax;
    xmin = ymin = 1e9, xmax = ymax = -1e9;
    for (int i = 1; i <= n; ++i) {
        double tx = x[i] + dx[d[i]] * now, ty = y[i] + dy[d[i]] * now;
        cmin(xmin, tx), cmax(xmax, tx), cmin(ymin, ty), cmax(ymax, ty);
    }
    return (xmax - xmin) * (ymax - ymin);
}

int main() {
#ifdef whyqx
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif
    cin >> n;
    char op[10];
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", x + i, y + i);
        scanf("%s", op);
        if (*op == 'R') d[i] = 0;
        if (*op == 'L') d[i] = 1;
        if (*op == 'U') d[i] = 2;
        if (*op == 'D') d[i] = 3;
    }
    double l = 0, r = 1e18;
    for (int i = 1; i <= 500; ++i) {
        double tmp = (r - l) / 3.0, midl = l + tmp, midr = r - tmp, resl, resr;
        resl = get(midl), resr = get(midr);
        if (resl < resr) r = midr;
        else l = midl;
    }
    printf("%.11f\n", get(l));
    return 0;
}

### Prompt

Create a solution in cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define N 100005
using namespace std;

template <typename T> inline void cmax(T& x, T y) {
    if (x < y) x = y;
}

template <typename T> inline void cmin(T& x, T y) {
    if (x > y) x = y;
}

const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
int n, x[N], y[N], d[N];

double get(double now) {
    double xmin, xmax, ymin, ymax;
    xmin = ymin = 1e9, xmax = ymax = -1e9;
    for (int i = 1; i <= n; ++i) {
        double tx = x[i] + dx[d[i]] * now, ty = y[i] + dy[d[i]] * now;
        cmin(xmin, tx), cmax(xmax, tx), cmin(ymin, ty), cmax(ymax, ty);
    }
    return (xmax - xmin) * (ymax - ymin);
}

int main() {
#ifdef whyqx
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif
    cin >> n;
    char op[10];
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", x + i, y + i);
        scanf("%s", op);
        if (*op == 'R') d[i] = 0;
        if (*op == 'L') d[i] = 1;
        if (*op == 'U') d[i] = 2;
        if (*op == 'D') d[i] = 3;
    }
    double l = 0, r = 1e18;
    for (int i = 1; i <= 500; ++i) {
        double tmp = (r - l) / 3.0, midl = l + tmp, midr = r - tmp, resl, resr;
        resl = get(midl), resr = get(midr);
        if (resl < resr) r = midr;
        else l = midl;
    }
    printf("%.11f\n", get(l));
    return 0;
}

```