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>
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
template <class T>
inline void read(T& x) {
  bool sign = 0;
  char c = getchar();
  while (c != '-' && (c > '9' || c < '0')) c = getchar();
  if (c == '-')
    x = 0, sign = 1;
  else
    x = c - '0';
  while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
  if (sign) x *= -1;
}
using namespace std;
const int maxn = 2e5 + 10;
char s[maxn];
long long prex[maxn], prey[maxn];
int main() {
  int t;
  for (read(t); t--;) {
    scanf("%s", s + 1);
    int len = 0;
    int mx[2][2] = {0, 0, 0, 0}, my[2][2] = {0, 0, 0, 0};
    for (int i = 1; s[i]; ++i, ++len) {
      prex[i] = prey[i] = 0;
      switch (s[i]) {
        case 'W':
          prex[i] = 1;
          break;
        case 'S':
          prex[i] = -1;
          break;
        case 'D':
          prey[i] = 1;
          break;
        case 'A':
          prey[i] = -1;
          break;
      }
      prex[i] += prex[i - 1];
      prey[i] += prey[i - 1];
      if (prex[i] < prex[mx[0][0]])
        mx[0][0] = mx[0][1] = i;
      else if (prex[i] == prex[mx[0][1]])
        mx[0][1] = i;
      if (prex[i] > prex[mx[1][0]])
        mx[1][0] = mx[1][1] = i;
      else if (prex[i] == prex[mx[1][1]])
        mx[1][1] = i;
      if (prey[i] < prey[my[0][0]])
        my[0][0] = my[0][1] = i;
      else if (prey[i] == prey[my[0][1]])
        my[0][1] = i;
      if (prey[i] > prey[my[1][0]])
        my[1][0] = my[1][1] = i;
      else if (prey[i] == prey[my[1][1]])
        my[1][1] = i;
    }
    long long res = (prex[mx[1][0]] - prex[mx[0][0]] + 1ll) *
                    (prey[my[1][0]] - prey[my[0][0]] + 1ll);
    if (mx[1][0] > mx[0][1] || mx[1][1] < mx[0][0])
      res = min(res, max(2ll, (prex[mx[1][0]] - prex[mx[0][0]])) *
                         (prey[my[1][0]] - prey[my[0][0]] + 1ll));
    if (my[1][0] > my[0][1] || my[1][1] < my[0][0])
      res = min(res, (prex[mx[1][0]] - prex[mx[0][0]] + 1ll) *
                         max(2ll, (prey[my[1][0]] - prey[my[0][0]])));
    printf("%lld\n", res);
  }
}
