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;
int t;
string s;
int xx[300111], yy[300111];
int minx[300111], maxx[300111], miny[300111], maxy[300111];
int main() {
  cin >> t;
  while (t--) {
    int x = 0, y = 0;
    cin >> s;
    if (s.size() == 1) {
      cout << "2\n";
      continue;
    }
    int xmin = 0, ymin = 0, xmax = 0, ymax = 0;
    xx[0] = yy[0] = 0;
    for (int i = 0; i < s.size(); ++i) {
      switch (s[i]) {
        case 'W':
          ++y;
          ymin = min(ymin, y);
          ymax = max(ymax, y);
          break;
        case 'A':
          --x;
          xmin = min(xmin, x);
          xmax = max(xmax, x);
          break;
        case 'S':
          --y;
          ymin = min(ymin, y);
          ymax = max(ymax, y);
          break;
        case 'D':
          ++x;
          xmin = min(xmin, x);
          xmax = max(xmax, x);
          break;
      }
      xx[i + 1] = x;
      yy[i + 1] = y;
    }
    int nm = s.size() + 1;
    minx[nm] = 999999999;
    maxx[nm] = -999999999;
    miny[nm] = 999999999;
    maxy[nm] = -999999999;
    for (int i = s.size(); i >= 0; --i) {
      minx[i] = minx[i + 1];
      maxx[i] = maxx[i + 1];
      miny[i] = miny[i + 1];
      maxy[i] = maxy[i + 1];
      minx[i] = min(minx[i], xx[i]);
      maxx[i] = max(maxx[i], xx[i]);
      miny[i] = min(miny[i], yy[i]);
      maxy[i] = max(maxy[i], yy[i]);
    }
    minx[nm] = x;
    maxx[nm] = x;
    miny[nm] = y;
    maxy[nm] = y;
    long long w = xmax - xmin + 1;
    long long h = ymax - ymin + 1;
    long long ans = w * h;
    bool ok1 = false, ok2 = false, ok3 = false, ok4 = false;
    for (int i = 0; i <= s.size(); ++i)
      if (xx[i] == xmax) {
        if (minx[i + 1] != xmin) ok1 = true;
        break;
      }
    for (int i = 0; i <= s.size(); ++i)
      if (xx[i] == xmin) {
        if (maxx[i + 1] != xmax) ok3 = true;
        break;
      }
    for (int i = 0; i <= s.size(); ++i)
      if (yy[i] == ymax) {
        if (miny[i + 1] != ymin) ok2 = true;
        break;
      }
    for (int i = 0; i <= s.size(); ++i)
      if (yy[i] == ymin) {
        if (maxy[i + 1] != ymax) ok4 = true;
        break;
      }
    if ((ok1 || ok3) && w > 2) ans = min(ans, 1LL * max(1LL, w - 1) * h);
    if ((ok2 || ok4) && h > 2) ans = min(ans, 1LL * max(1LL, h - 1) * w);
    cout << ans << '\n';
  }
  return 0;
}
