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 sys

def calc(line):
    left = 0
    right = 0
    up = 0
    bottom = 0

    ls = [-1]
    rs = [-1]
    us = [-1]
    bs = [-1]

    clr = 0
    cub = 0

    i = 0
    for c in line:
        if c == 'A':
            clr -= 1
            if clr < left:
                left = clr
                ls = [i]
            elif clr == left:
                ls.append(i)
        elif c == 'D':
            clr += 1
            if clr > right:
                right = clr
                rs = [i]
            elif clr == right:
                rs.append(i)
        elif c == 'W':
            cub += 1
            if cub > up:
                up = cub
                us = [i]
            elif cub == up:
                us.append(i)
        elif c == 'S':
            cub -= 1
            if cub < bottom:
                bottom = cub
                bs=[i]
            elif cub == bottom:
                bs.append(i)
        i+=1
    a = up-bottom+1
    b = right - left+1

    if up - bottom < right - left:
        if can(us,bs,a):
            a -= 1
        elif can(ls, rs,b):
            b -= 1
    else:
        if can(ls, rs,b):
            b -= 1
        elif can(us,bs,a):
            a -= 1
    return a * b

def can(s,b,n):
    if n <= 2:
        return False
    maxs = max(s)
    mins = min(s)
    maxb = max(b)
    minb = min(b)
    if maxs < minb or mins > maxb:
        return True
    else:
        return False

n = int(sys.stdin.readline().strip())
for i in range(n):
    line = sys.stdin.readline().strip()
    r = calc(line)
    print r
	    		  			 	  	    	 			  			