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;
string ip;
int x, y;
int mxUp[200050];
int mxDown[200050];
int mxLeft[200050];
int mxRight[200050];
int remX[200050];
int remY[200050];
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> ip;
    x = y = 0;
    int s = ip.size();
    for (int i = 1; i <= s; ++i) {
      switch (ip[i - 1]) {
        case 'W':
          y++;
          break;
        case 'S':
          y--;
          break;
        case 'A':
          x--;
          break;
        case 'D':
          x++;
      }
      remY[i] = mxUp[i] = mxDown[i] = y;
      remX[i] = mxLeft[i] = mxRight[i] = x;
    }
    mxUp[0] = mxDown[0] = mxLeft[0] = mxRight[0] = remX[0] = remY[0] = 0;
    for (int i = s - 1; i >= 0; --i) {
      mxUp[i] = max(mxUp[i + 1], mxUp[i]);
      mxDown[i] = min(mxDown[i + 1], mxDown[i]);
      mxLeft[i] = min(mxLeft[i + 1], mxLeft[i]);
      mxRight[i] = max(mxRight[i + 1], mxRight[i]);
    }
    long long ud = mxUp[0] - mxDown[0] + 1;
    long long lr = mxRight[0] - mxLeft[0] + 1;
    long long squre = ud * lr;
    int mxX = 0, mnX = 0, mnY = 0, mxY = 0;
    for (int i = 1; i < s; ++i) {
      mxX = max(remX[i], mxX);
      mnX = min(remX[i], mnX);
      mxY = max(remY[i], mxY);
      mnY = min(remY[i], mnY);
      squre = min(
          squre, (max(mxX, mxRight[i] - 1) - min(mnX, mxLeft[i] - 1) + 1) * ud);
      squre = min(
          squre, (max(mxX, mxRight[i] + 1) - min(mnX, mxLeft[i] + 1) + 1) * ud);
      squre = min(squre,
                  (max(mxY, mxUp[i] - 1) - min(mnY, mxDown[i] - 1) + 1) * lr);
      squre = min(squre,
                  (max(mxY, mxUp[i] + 1) - min(mnY, mxDown[i] + 1) + 1) * lr);
    }
    cout << squre << endl;
  }
}
