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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 5e5 + 5, M = 2 * N + 5;
int n, t;
char second[N];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%s", &second);
    n = strlen(second);
    long long l = 0, r = 0, lmx = 0, lmn = 0, rmx = 0, rmn = 0;
    long long lmxlen = 0, lmnlen = 0, rmxlen = 0, rmnlen = 0;
    for (int i = 0; i < n; ++i) {
      if (second[i] == 'W')
        lmx = max(lmx, ++l);
      else if (second[i] == 'S')
        lmn = min(lmn, --l);
      else if (second[i] == 'D')
        rmx = max(rmx, ++r);
      else
        rmn = min(rmn, --r);
      lmxlen = max(lmxlen, lmx - l);
      lmnlen = max(lmnlen, l - lmn);
      rmxlen = max(rmxlen, rmx - r);
      rmnlen = max(rmnlen, r - rmn);
    }
    long long ans = (lmx - lmn + 1LL) * (rmx - rmn + 1LL);
    if (lmx - lmn > 1LL && lmxlen != lmnlen)
      ans = min(ans, (lmx - lmn) * (rmx - rmn + 1LL));
    if (rmx - rmn > 1LL && rmxlen != rmnlen)
      ans = min(ans, (lmx - lmn + 1LL) * (rmx - rmn));
    printf("%lld\n", ans);
  }
  return 0;
}
