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 = 2000005;
const int mod = 998244353;
char s[maxn];
long long maxlast, maxfirst, minlast, minfirst;
long long solve(char a, char b) {
  int len = strlen(s);
  maxlast = maxfirst = minlast = minfirst = 0;
  long long y = 0, mx = 0, mi = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] == a) y++;
    if (s[i] == b) y--;
    if (mx < y) {
      mx = y;
      maxlast = maxfirst = i;
    }
    if (mx == y) maxlast = i;
    if (mi > y) {
      mi = y;
      minlast = minfirst = i;
    }
    if (mi == y) minlast = i;
  }
  return mx - mi + 1;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s);
    int f1 = 0, f2 = 0;
    long long n = solve('W', 'S');
    if ((maxlast < minfirst || minlast < maxfirst) && n > 2) f1 = 1;
    long long m = solve('A', 'D');
    if ((maxlast < minfirst || minlast < maxfirst) && m > 2) f2 = 1;
    long long ans = n * m;
    if (f1 && f2)
      ans = min(n * (m - 1), m * (n - 1));
    else if (f1)
      ans = m * (n - 1);
    else if (f2)
      ans = n * (m - 1);
    printf("%lld\n", ans);
  }
  return 0;
}
