#include <algorithm>
#include <bitset>
#include <cfloat>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits.h>
#include <list>
#include <map>
#include <math.h>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define int long long
#define ll long long
#define eps DBL_EPSILON
#define mod (ll)998244353
#define INF LLONG_MAX/10
#define P pair<int,int>
#define prique priority_queue
using namespace std;
int n, x[100010], y[100010];
char d[100010];
long double func(long double time) {
	long double mi = 200000000, ma = -200000000;
	rep(i, n) {
		if (d[i] == 'R') {
			mi = min(mi, x[i] + time);
			ma = max(ma, x[i] + time);
		}
		else if (d[i] == 'L') {
			mi = min(mi, x[i] - time);
			ma = max(ma, x[i] - time);
		}
		else {
			mi = min(mi, (long double)x[i]);
			ma = max(ma, (long double)x[i]);
		}
	}
	long double memo = ma - mi;
	mi = 200000000; ma = -200000000;
	rep(i, n) {
		if (d[i] == 'U') {
			mi = min(mi, y[i] + time);
			ma = max(ma, y[i] + time);
		}
		else if (d[i] == 'D') {
			mi = min(mi, y[i] - time);
			ma = max(ma, y[i] - time);
		}
		else {
			mi = min(mi, (long double)y[i]);
			ma = max(ma, (long double)y[i]);
		}
	}
	return memo * (ma - mi);
}
signed main() {
	cin >> n;
	rep(i, n)cin >> x[i] >> y[i] >> d[i];
	long double ok = 0, ng = 300000000;
	rep(i, 500) {
		long double md1 = ok + (ng - ok) / 3, md2 = ok + (ng - ok) * 2 / 3;
		if (func(md1) <= func(md2))ng = md2;
		else ok = md1;
	}
	printf("%.12Lf\n", func(ng));
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <algorithm>
#include <bitset>
#include <cfloat>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits.h>
#include <list>
#include <map>
#include <math.h>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define int long long
#define ll long long
#define eps DBL_EPSILON
#define mod (ll)998244353
#define INF LLONG_MAX/10
#define P pair<int,int>
#define prique priority_queue
using namespace std;
int n, x[100010], y[100010];
char d[100010];
long double func(long double time) {
	long double mi = 200000000, ma = -200000000;
	rep(i, n) {
		if (d[i] == 'R') {
			mi = min(mi, x[i] + time);
			ma = max(ma, x[i] + time);
		}
		else if (d[i] == 'L') {
			mi = min(mi, x[i] - time);
			ma = max(ma, x[i] - time);
		}
		else {
			mi = min(mi, (long double)x[i]);
			ma = max(ma, (long double)x[i]);
		}
	}
	long double memo = ma - mi;
	mi = 200000000; ma = -200000000;
	rep(i, n) {
		if (d[i] == 'U') {
			mi = min(mi, y[i] + time);
			ma = max(ma, y[i] + time);
		}
		else if (d[i] == 'D') {
			mi = min(mi, y[i] - time);
			ma = max(ma, y[i] - time);
		}
		else {
			mi = min(mi, (long double)y[i]);
			ma = max(ma, (long double)y[i]);
		}
	}
	return memo * (ma - mi);
}
signed main() {
	cin >> n;
	rep(i, n)cin >> x[i] >> y[i] >> d[i];
	long double ok = 0, ng = 300000000;
	rep(i, 500) {
		long double md1 = ok + (ng - ok) / 3, md2 = ok + (ng - ok) * 2 / 3;
		if (func(md1) <= func(md2))ng = md2;
		else ok = md1;
	}
	printf("%.12Lf\n", func(ng));
	return 0;
}
```