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).
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 22;
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    long long minL, minR, maxL, maxR;
    long long lastA, lastD, lastS, lastW;
    long long newA, newS, newD, newW;
    newA = newS = newD = newW = 0;
    lastA = lastD = lastS = lastW = 0;
    minL = minR = maxL = maxR = 0;
    long long x = 0, y = 0;
    for (long long i = 0; i < (s.size()); i++) {
      if (s[i] == 'A')
        x--;
      else if (s[i] == 'D')
        x++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'W')
        y++;
      if (minL > x) newA = i;
      if (maxL < x) newD = i;
      if (minR > y) newS = i;
      if (maxR < y) newW = i;
      minL = min(minL, x);
      maxL = max(maxL, x);
      minR = min(minR, y);
      maxR = max(maxR, y);
      if (minL == x) lastA = i;
      if (maxL == x) lastD = i;
      if (minR == y) lastS = i;
      if (maxR == y) lastW = i;
    }
    long long height = (maxR - minR + 1), width = (maxL - minL + 1);
    long long res = height * width;
    long long dec = 0;
    if (newW > lastS && height != 2) dec = max(dec, width);
    if (newS > lastW && height != 2) dec = max(dec, width);
    if (newA > lastD && width != 2) dec = max(dec, height);
    if (newD > lastA && width != 2) dec = max(dec, height);
    cout << res - dec << endl;
  }
  return 0;
}
