You have a string s — a sequence of commands for your toy robot. The robot is placed in some cell of a rectangular grid. He can perform four commands:

  * 'W' — move one cell up; 
  * 'S' — move one cell down; 
  * 'A' — move one cell left; 
  * 'D' — move one cell right. 



Let Grid(s) be the grid of minimum possible area such that there is a position in the grid where you can place the robot in such a way that it will not fall from the grid while running the sequence of commands s. For example, if s = DSAWWAW then Grid(s) is the 4 × 3 grid:

  1. you can place the robot in the cell (3, 2); 
  2. the robot performs the command 'D' and moves to (3, 3); 
  3. the robot performs the command 'S' and moves to (4, 3); 
  4. the robot performs the command 'A' and moves to (4, 2); 
  5. the robot performs the command 'W' and moves to (3, 2); 
  6. the robot performs the command 'W' and moves to (2, 2); 
  7. the robot performs the command 'A' and moves to (2, 1); 
  8. the robot performs the command 'W' and moves to (1, 1). 

<image>

You have 4 extra letters: one 'W', one 'A', one 'S', one 'D'. You'd like to insert at most one of these letters in any position of sequence s to minimize the area of Grid(s).

What is the minimum area of Grid(s) you can achieve?

Input

The first line contains one integer T (1 ≤ T ≤ 1000) — the number of queries.

Next T lines contain queries: one per line. This line contains single string s (1 ≤ |s| ≤ 2 ⋅ 10^5, s_i ∈ \{W, A, S, D\}) — the sequence of commands.

It's guaranteed that the total length of s over all queries doesn't exceed 2 ⋅ 10^5.

Output

Print T integers: one per query. For each query print the minimum area of Grid(s) you can achieve.

Example

Input


3
DSAWWAW
D
WA


Output


8
2
4

Note

In the first query you have to get string DSAWW\underline{D}AW.

In second and third queries you can not decrease the area of Grid(s).
import java.util.*;
import java.lang.*;

public class solution {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t, current_x, current_y, up_max, down_max, left_max, right_max, first_up, last_up, first_down, last_down, first_left, last_left, first_right, last_right, horizontal_count, vertical_count;
		long h, w;
		String s;
		
		t = sc.nextInt();
		for (int c=0;c<t;++c){
			s = sc.next();
			up_max = down_max = right_max = left_max = 0;
			first_up = last_up = first_down = last_down = first_left = last_left = first_right = last_right = 0;
			current_x = current_y = 0;
			horizontal_count = vertical_count = 0;
			for (int i = 0; i < s.length(); i++) {
				if (s.charAt(i) == 'W') {
					++current_y;
					++vertical_count;
					if (current_y > up_max) {
						up_max = current_y;
						first_up = last_up = i + 1;
					}
					else if (current_y == up_max)
						last_up = i + 1;
				}
				else if (s.charAt(i) == 'S') {
					--current_y;
					++vertical_count;
					if (current_y < down_max) {
						down_max = current_y;
						first_down = last_down = i + 1;
					}
					else if (current_y == down_max)
						last_down = i + 1;
				}
				else if (s.charAt(i) == 'D') {
					++current_x;
					++horizontal_count;
					if (current_x > right_max) {
						right_max = current_x;
						first_right = last_right = i + 1;
					}
					else if (current_x == right_max)
						last_right = i + 1;
				}
				else {
					--current_x;
					++horizontal_count;
					if (current_x < left_max) {
						left_max = current_x;
						first_left = last_left = i + 1;
					}
					else if (current_x == left_max)
						last_left = i + 1;
				}
			}
			h = up_max - down_max + 1;
			w = right_max - left_max + 1;
			long ans = h*w;
			if (vertical_count > 1 && last_up < first_down)
				ans = Math.min(ans, (h - 1)*w);
			if (vertical_count > 1 && last_down < first_up)
				ans = Math.min(ans, (h - 1)*w);
			if (horizontal_count > 1 && last_right < first_left)
				ans = Math.min(ans, h*(w - 1));
			if (horizontal_count > 1 && last_left < first_right)
				ans = Math.min(ans, h*(w - 1));

			System.out.println(ans);
		}	
	}
}
