/*
[abc130] F - Minimum Bounding Box
*/

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;

const int MAX_N = 1e5;
const int MAX_XY = 1e8;
const double INF = 1e18;

int dxy[2][4] = {{0, 1, 0, -1}, {1, 0, -1, 0}};

int N;
int xy[2][MAX_N];
int d[MAX_N];

double calc(double t) {
    double m[2][2];
    for (int c = 0; c < 2; c++) {
        m[c][0] = INF;
        m[c][1] = -INF;
        for (int i = 0; i < N; i++) {
            m[c][0] = min(m[c][0], xy[c][i] + dxy[c][d[i]] * t);
            m[c][1] = max(m[c][1], xy[c][i] + dxy[c][d[i]] * t);
        }
    }

    return (m[0][1] - m[0][0]) * (m[1][1] - m[1][0]);
}

double solve() {
    double l = 0, r = 2 * MAX_XY + 1;
    for (int i = 0; i < 200; i++) {
        double m1 = (2 * l + r) / 3.0, m2 = (l + 2 * r) / 3.0;
        if (calc(m1) < calc(m2)) {
            r = m2;
        } else {
            l = m1;
        }
    }

    return calc(l);
}

int main() {
    cin >> N;
    for (int i = 0; i < N; i++) {
        char d_raw;
        cin >> xy[0][i] >> xy[1][i] >> d_raw;
        switch (d_raw) {
            case 'U':
                d[i] = 0;
                break;
            case 'R':
                d[i] = 1;
                break;
            case 'D':
                d[i] = 2;
                break;
            case 'L':
                d[i] = 3;
                break;
        }
    }

    cout << setprecision(16) << fixed << solve() << endl;

    return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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
/*
[abc130] F - Minimum Bounding Box
*/

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;

const int MAX_N = 1e5;
const int MAX_XY = 1e8;
const double INF = 1e18;

int dxy[2][4] = {{0, 1, 0, -1}, {1, 0, -1, 0}};

int N;
int xy[2][MAX_N];
int d[MAX_N];

double calc(double t) {
    double m[2][2];
    for (int c = 0; c < 2; c++) {
        m[c][0] = INF;
        m[c][1] = -INF;
        for (int i = 0; i < N; i++) {
            m[c][0] = min(m[c][0], xy[c][i] + dxy[c][d[i]] * t);
            m[c][1] = max(m[c][1], xy[c][i] + dxy[c][d[i]] * t);
        }
    }

    return (m[0][1] - m[0][0]) * (m[1][1] - m[1][0]);
}

double solve() {
    double l = 0, r = 2 * MAX_XY + 1;
    for (int i = 0; i < 200; i++) {
        double m1 = (2 * l + r) / 3.0, m2 = (l + 2 * r) / 3.0;
        if (calc(m1) < calc(m2)) {
            r = m2;
        } else {
            l = m1;
        }
    }

    return calc(l);
}

int main() {
    cin >> N;
    for (int i = 0; i < N; i++) {
        char d_raw;
        cin >> xy[0][i] >> xy[1][i] >> d_raw;
        switch (d_raw) {
            case 'U':
                d[i] = 0;
                break;
            case 'R':
                d[i] = 1;
                break;
            case 'D':
                d[i] = 2;
                break;
            case 'L':
                d[i] = 3;
                break;
        }
    }

    cout << setprecision(16) << fixed << solve() << endl;

    return 0;
}

```