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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int MAXN = 200005;
char s[MAXN];
int n, m, f[MAXN], g[MAXN];
int main() {
  int T = read();
  while (T--) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    int a = 0, b = 0, ma = 0, Ma = 0, mb = 0, Mb = 0;
    for (int i = (1); i <= (n); ++i) {
      if (s[i] == 'W')
        a++;
      else if (s[i] == 'S')
        a--;
      if (s[i] == 'A')
        b--;
      else if (s[i] == 'D')
        b++;
      f[i] = a, g[i] = b;
      if (f[i] < f[ma]) ma = i;
      if (f[i] > f[Ma]) Ma = i;
      if (g[i] < g[mb]) mb = i;
      if (g[i] > g[Mb]) Mb = i;
    }
    long long ans = 1ll * (f[Ma] - f[ma] + 1) * (g[Mb] - g[mb] + 1);
    int l1 = n + 1, r1 = 0, l2 = n + 1, r2 = 0;
    for (int i = (0); i <= (n); ++i) {
      if (f[i] == f[ma]) l1 = min(l1, i), r1 = max(r1, i);
      if (f[i] == f[Ma]) l2 = min(l2, i), r2 = max(r2, i);
    }
    if (f[Ma] - f[ma] > 1 && (r1 < l2 || r2 < l1))
      ans = min(ans, 1ll * (f[Ma] - f[ma]) * (g[Mb] - g[mb] + 1));
    l1 = n + 1, r1 = 0, l2 = n + 1, r2 = 0;
    for (int i = (0); i <= (n); ++i) {
      if (g[i] == g[mb]) l1 = min(l1, i), r1 = max(r1, i);
      if (g[i] == g[Mb]) l2 = min(l2, i), r2 = max(r2, i);
    }
    if (g[Mb] - g[mb] > 1 && (r1 < l2 || r2 < l1))
      ans = min(ans, 1ll * (f[Ma] - f[ma] + 1) * (g[Mb] - g[mb]));
    printf("%lld\n", ans);
  }
  return 0;
}
