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 main() {
  int q;
  cin >> q;
  long long int results[q];
  for (int j = 0; j < q; j++) {
    long long int leftSway = 0, leftMax = 0, rightMax = 0, upSway = 0,
                  upMax = 0, downMax = 0;
    string s;
    cin >> s;
    int a = INT_MAX, b = INT_MAX;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'A') leftSway++;
      if (s[i] == 'D') leftSway--;
      leftMax = max(leftMax, leftSway);
      rightMax = min(rightMax, leftSway);
      if (s[i] == 'W') upSway++;
      if (s[i] == 'S') upSway--;
      upMax = max(upMax, upSway);
      downMax = min(downMax, upSway);
    }
    results[j] = (leftMax - rightMax + 1) * (upMax - downMax + 1);
    if (leftMax - rightMax + 1 >= 3) {
      int firstMin = -1, lastMin = -1, firstMax = -1, lastMax = -1;
      int l = 0;
      for (int i = 0; i < s.size(); i++) {
        if (s[i] == 'A') l++;
        if (s[i] == 'D') l--;
        if (firstMin == -1 && rightMax == l) {
          firstMin = i;
        }
        if (rightMax == l) {
          lastMin = i;
        }
        if (firstMax == -1 && leftMax == l) {
          firstMax = i;
        }
        if (leftMax == l) {
          lastMax = i;
        }
      }
      if (firstMax > lastMin && leftMax != 0) {
        results[j] = (leftMax - rightMax) * (upMax - downMax + 1);
      }
      if (lastMax < firstMin && rightMax != 0) {
        results[j] = (leftMax - rightMax) * (upMax - downMax + 1);
      }
    }
    if (upMax - downMax + 1 >= 3) {
      int firstMin = -1, lastMin = -1, firstMax = -1, lastMax = -1;
      int l = 0, u = 0;
      for (int i = 0; i < s.size(); i++) {
        if (s[i] == 'W') u++;
        if (s[i] == 'S') u--;
        if (firstMin == -1 && downMax == u) {
          firstMin = i;
        }
        if (downMax == u) {
          lastMin = i;
        }
        if (firstMax == -1 && upMax == u) {
          firstMax = i;
        }
        if (upMax == u) {
          lastMax = i;
        }
      }
      if (firstMax > lastMin && upMax != 0) {
        results[j] =
            min(results[j], (leftMax - rightMax + 1) * (upMax - downMax));
      }
      if (lastMax < firstMin && downMax != 0) {
        results[j] =
            min(results[j], (leftMax - rightMax + 1) * (upMax - downMax));
      }
    }
  }
  for (int i = 0; i < q; i++) cout << results[i] << endl;
  return 0;
}
