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;
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      ans = ((ans) * (a));
    }
    a = ((a) * (a));
    b = b / 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int l = 0, r = 0, t = 0, d = 0;
    long long int lenl = 0, lenr = 0, lent = 0, lend = 0;
    long long int x = 0, y = 0;
    for (long long int i = 0; i < s.length(); i++) {
      if (s[i] == 'W') {
        y++;
        t = max(t, y);
      } else if (s[i] == 'S') {
        y--;
        d = min(d, y);
      } else if (s[i] == 'A') {
        x--;
        l = min(x, l);
      } else {
        x++;
        r = max(x, r);
      }
      lenl = max(lenl, x - l);
      lenr = max(lenr, r - x);
      lent = max(lent, t - y);
      lend = max(lend, y - d);
    }
    long long int ans = (r - l + 1) * (t - d + 1);
    if ((r - l) > 1 && lenl != lenr) {
      ans = min(ans, (r - l) * (t - d + 1));
    }
    if ((t - d) > 1 && lent != lend) {
      ans = min(ans, (t - d) * (r - l + 1));
    }
    cout << ans << endl;
  }
  return 0;
}
