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). 

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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).
def lim(s):
    now = 0
    up, down = 0, 0
    for i in s:
        now += i
        up = max(up, now)
        down = min(down, now)
    return up, down
def f(a):
    return a[0] - a[1] + 1
def upg(s):
    t = lim(s)
    up, down = t[0], t[1]
    arr = [1, 1]
    now = 0
    for i in range(len(s) - 1):
        if now == up - 1 and s[i + 1] == 1 and arr[0] == 1:
            arr[0] = 0
            if f(lim(s[:(i + 1)] + [-1] + s[(i + 1):])) < f(t):
                return 1
        if now == down + 1 and s[i + 1] == -1 and arr[1] == 1:
            arr[1] = 0
            if f(lim(s[:(i + 1)] + [1] + s[(i + 1):])) < f(t):
                return 1
        now += s[i + 1]
    return 0


for q in range(int(input())):
    s = input()
    s1, s2 = [0], [0]
    for i in s:
        if i == 'W': s1.append(1)
        if i == 'S': s1.append(-1)
        if i == 'A': s2.append(1)
        if i == 'D': s2.append(-1)
    u1 = upg(s1)
    u2 = upg(s2)
    res1, res2 = f(lim(s1)), f(lim(s2))
    ans = min((res1 - u1) * res2, (res2 - u2) * res1)
    print(ans)