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 read() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long N = 2e5 + 5;
char s[N];
long long a[N], b[N], sum1[N], sum2[N], mi1[N], mx1[N], mi2[N], mx2[N], n, num,
    cnt;
signed main() {
  long long T = read();
  while (T--) {
    scanf("%s", s + 1);
    n = strlen(s + 1), num = 0, cnt = 0;
    for (long long i = 1; i <= n; ++i) {
      if (s[i] == 'W') a[++num] = 1;
      if (s[i] == 'S') a[++num] = -1;
      if (s[i] == 'A') b[++cnt] = -1;
      if (s[i] == 'D') b[++cnt] = 1;
    }
    for (long long i = 1; i <= num; ++i)
      sum1[i] = a[i] + sum1[i - 1], mi1[i] = min(sum1[i], mi1[i - 1]),
      mx1[i] = max(sum1[i], mx1[i - 1]);
    for (long long i = 1; i <= cnt; ++i)
      sum2[i] = b[i] + sum2[i - 1], mi2[i] = min(sum2[i], mi2[i - 1]),
      mx2[i] = max(sum2[i], mx2[i - 1]);
    long long Ans = (mx1[num] - mi1[num] + 1) * (mx2[cnt] - mi2[cnt] + 1);
    long long x = mx1[num] - mi1[num], y = mx2[cnt] - mi2[cnt];
    long long mx = 0, mi = 0, Mi, Mx, xx = x, yy = y;
    for (long long i = num; i >= 1; --i) {
      mi = min(sum1[i] - 1, mi), mx = max(sum1[i] - 1, mx);
      Mi = min(mi, mi1[i]), Mx = max(mx, mx1[i]);
      xx = min(xx, Mx - Mi);
    }
    Ans = min(Ans, (xx + 1) * (y + 1));
    xx = x, mx = 0, mi = 0;
    for (long long i = num; i >= 1; --i) {
      mi = min(sum1[i] + 1, mi), mx = max(sum1[i] + 1, mx);
      Mi = min(mi, mi1[i]), Mx = max(mx, mx1[i]);
      xx = min(xx, Mx - Mi);
    }
    Ans = min(Ans, (xx + 1) * (y + 1));
    mx = mi = 0;
    yy = y;
    for (long long i = cnt; i >= 1; --i) {
      mi = min(sum2[i] - 1, mi), mx = max(sum2[i] - 1, mx);
      Mi = min(mi, mi2[i]), Mx = max(mx, mx2[i]);
      yy = min(yy, Mx - Mi);
    }
    Ans = min(Ans, (x + 1) * (yy + 1));
    mx = mi = 0;
    yy = y;
    for (long long i = cnt; i >= 1; --i) {
      mi = min(sum2[i] + 1, mi), mx = max(sum2[i] + 1, mx);
      Mi = min(mi, mi2[i]), Mx = max(mx, mx2[i]);
      yy = min(yy, Mx - Mi);
    }
    Ans = min(Ans, (x + 1) * (yy + 1));
    printf("%I64d\n", Ans);
  }
  return 0;
}
