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 int n = 2e5 + 10;
int T;
string s;
long long x[n], y[n];
long long mxX[n], mxY[n], revY[n], revX[n];
long long miX[n], miY[n], reviY[n], reviX[n];
int X[4] = {-1, 0, 1, 0};
int Y[4] = {0, -1, 0, 1};
long long area(long long xmi, long long xma, long long ymi, long long yma) {
  return (xma - xmi + 1) * (yma - ymi + 1);
}
int main() {
  cin >> T;
  while (T--) {
    cin >> s;
    for (int i = 0; i < ((int)(s).size()); i++) {
      mxX[i + 1] = mxY[i + 1] = revY[i + 1] = revX[i + 1] = 0;
      x[i + 1] = y[i + 1] = 0;
      switch (s[i]) {
        case 'W':
          y[i + 1] = 1;
          break;
        case 'S':
          y[i + 1] = -1;
          break;
        case 'A':
          x[i + 1] = -1;
          break;
        case 'D':
          x[i + 1] = 1;
          break;
      }
      x[i + 1] += x[i];
      y[i + 1] += y[i];
    }
    for (int i = 1; i <= ((int)(s).size()); i++) {
      mxX[i] = max(mxX[i - 1], x[i]);
      mxY[i] = max(mxY[i - 1], y[i]);
      miX[i] = min(miX[i - 1], x[i]);
      miY[i] = min(miY[i - 1], y[i]);
    }
    reviX[((int)(s).size())] = revX[((int)(s).size())] = x[((int)(s).size())];
    reviY[((int)(s).size())] = revY[((int)(s).size())] = y[((int)(s).size())];
    for (int i = ((int)(s).size()) - 1; i >= 0; i--) {
      revX[i] = max(revX[i + 1], x[i]);
      revY[i] = max(revY[i + 1], y[i]);
      reviX[i] = min(reviX[i + 1], x[i]);
      reviY[i] = min(reviY[i + 1], y[i]);
    }
    long long best = area(miX[((int)(s).size())], mxX[((int)(s).size())],
                          miY[((int)(s).size())], mxY[((int)(s).size())]);
    for (int i = 0; i < ((int)(s).size()); i++) {
      for (int j = 0; j < 4; j++) {
        int dx = X[j];
        int dy = Y[j];
        best = min(best,
                   area(min(miX[i], reviX[i] + dx), max(mxX[i], revX[i] + dx),
                        min(miY[i], reviY[i] + dy), max(mxY[i], revY[i] + dy)));
      }
    }
    cout << best << "\n";
  }
  return 0;
}
