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 maxn = 200005;
int G, n, L, R, U, D, x, y, mx, my, stepx[maxn], stepy[maxn];
bool canX, canY;
char S[maxn];
int main() {
  scanf("%d", &G);
  while (G--) {
    scanf("%s", S);
    n = strlen(S);
    L = R = U = D = x = y = mx = my = 0;
    for (int i = 0; i < n; i++)
      if (S[i] == 'A')
        x--, L = min(L, x), stepx[++mx] = x;
      else if (S[i] == 'D')
        x++, R = max(R, x), stepx[++mx] = x;
      else if (S[i] == 'S')
        y--, D = min(D, y), stepy[++my] = y;
      else
        y++, U = max(U, y), stepy[++my] = y;
    canX = canY = false;
    if (R)
      for (int i = 1; i <= mx; i++)
        if (stepx[i] == R) {
          bool boo = true;
          for (int j = i + 1; j <= mx; j++)
            if (stepx[j] - 1 < L) {
              boo = false;
              break;
            }
          if (boo) canX = true;
          break;
        }
    if (L && !canX)
      for (int i = 1; i <= mx; i++)
        if (stepx[i] == L) {
          bool boo = true;
          for (int j = i + 1; j <= mx; j++)
            if (R < stepx[j] + 1) {
              boo = false;
              break;
            }
          if (boo) canX = true;
          break;
        }
    if (U)
      for (int i = 1; i <= my; i++)
        if (stepy[i] == U) {
          bool boo = true;
          for (int j = i + 1; j <= my; j++)
            if (stepy[j] - 1 < D) {
              boo = false;
              break;
            }
          if (boo) canY = true;
          break;
        }
    if (D && !canY)
      for (int i = 1; i <= my; i++)
        if (stepy[i] == D) {
          bool boo = true;
          for (int j = i + 1; j <= my; j++)
            if (U < stepy[j] + 1) {
              boo = false;
              break;
            }
          if (boo) canY = true;
          break;
        }
    int X = R - L + 1, Y = U - D + 1;
    printf("%lld\n", min((long long)max(X - canX, 2) * Y,
                         (long long)X * max(Y - canY, 2)));
  }
  return 0;
}
