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 n;
  cin >> n;
  getchar();
  while (n--) {
    string s;
    int x, y, l, r, u, d, L, R, U, D;
    x = y = l = r = u = d = L = R = U = D = 0;
    getline(cin, s);
    for (int i = 0; i < s.size(); i++) {
      switch (s[i]) {
        case 'A':
          x--;
          if (x < l) {
            l = x;
            L = 1;
            R = 0;
          } else if (x == l)
            L = 1;
          break;
        case 'D':
          x++;
          if (x > r) {
            r = x;
            R = 1;
            L = 0;
          } else if (x == r)
            R = 1;
          break;
        case 'W':
          y++;
          if (y > u) {
            u = y;
            U = 1;
            D = 0;
          } else if (y == u)
            U = 1;
          break;
        case 'S':
          y--;
          if (y < d) {
            d = y;
            D = 1;
            U = 0;
          } else if (y == d)
            D = 1;
          break;
      }
    }
    x = r - l + 1;
    y = u - d + 1;
    long long mi = 1LL * x * y, k1 = mi, k2 = mi;
    if (x > 2 && L * R == 0) k1 = mi - y;
    if (y > 2 && U * D == 0) k2 = mi - x;
    cout << min(k1, k2) << endl;
  }
  return 0;
}
