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;
long long solve(string s) {
  long long maxx = 0, minx = 0, maxy = 0, miny = 0, x = 0, y = 0;
  for (auto i : s) {
    if (i == 'W')
      x++;
    else if (i == 'S')
      x--;
    else if (i == 'A')
      y--;
    else
      y++;
    maxx = max(maxx, x), minx = min(minx, x), maxy = max(maxy, y),
    miny = min(miny, y);
  }
  return (maxx - minx + 1) * (maxy - miny + 1);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    map<char, long long> m;
    long long maxx = 0, minx = 0, maxy = 0, miny = 0, x = 0, y = 0;
    cin >> s;
    for (long long j = 0; j < s.length(); j++) {
      char i = s[j];
      if (i == 'W')
        x++;
      else if (i == 'S')
        x--;
      else if (i == 'A')
        y--;
      else
        y++;
      if (maxx < x) maxx = x, m['W'] = j;
      if (minx > x) minx = x, m['S'] = j;
      if (maxy < y) maxy = y, m['D'] = j;
      if (miny > y) miny = y, m['A'] = j;
    }
    string t1 = s, t2 = s, t3 = s, t4 = s;
    cout << min({solve(s), solve(t1.insert(m['W'], "S")),
                 solve(t2.insert(m['A'], "D")), solve(t3.insert(m['D'], "A")),
                 solve(t4.insert(m['S'], "W"))})
         << '\n';
  }
}
