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;
char tab[200007];
int main() {
  long long int q, i, j, lew, pra, gor, dol, x, y, ilew, iles, iled, ilea, lstp,
      lstl, lstg, lstd;
  scanf("%lld", &q);
  for (j = 0; j < q; j++) {
    lew = 0;
    pra = 0;
    dol = 0;
    gor = 0;
    x = 0;
    y = 0;
    ilea = 0;
    ilew = 0;
    iles = 0;
    iled = 0;
    lstg = -1;
    lstd = -1;
    lstp = -1;
    lstl = -1;
    scanf("%s", tab);
    for (i = 0; i < strlen(tab); i++) {
      if (tab[i] == 'W') {
        y++;
        gor = max(gor, y);
      }
      if (tab[i] == 'D') {
        x++;
        pra = max(pra, x);
      }
      if (tab[i] == 'S') {
        y--;
        dol = min(dol, y);
      }
      if (tab[i] == 'A') {
        x--;
        lew = min(lew, x);
      }
    }
    x = 0;
    y = 0;
    for (i = 0; i < strlen(tab); i++) {
      if (y == gor) {
        if (lstg == -1) {
          lstg = i;
        }
        ilew = i;
      }
      if (y == dol) {
        if (lstd == -1) {
          lstd = i;
        }
        iles = i;
      }
      if (x == pra) {
        if (lstp == -1) {
          lstp = i;
        }
        iled = i;
      }
      if (x == lew) {
        if (lstl == -1) {
          lstl = i;
        }
        ilea = i;
      }
      if (tab[i] == 'W') {
        y++;
      }
      if (tab[i] == 'D') {
        x++;
      }
      if (tab[i] == 'S') {
        y--;
      }
      if (tab[i] == 'A') {
        x--;
      }
    }
    if (y == gor) {
      if (lstg == -1) {
        lstg = i;
      }
      ilew = i;
    }
    if (y == dol) {
      if (lstd == -1) {
        lstd = i;
      }
      iles = i;
    }
    if (x == pra) {
      if (lstp == -1) {
        lstp = i;
      }
      iled = i;
    }
    if (x == lew) {
      if (lstl == -1) {
        lstl = i;
      }
      ilea = i;
    }
    if ((gor - dol) > (pra - lew)) {
      if ((pra - lew >= 2) && ((lstl > iled) || (lstp > ilea))) {
        printf("%lld\n", (pra - lew) * (gor - dol + 1));
        continue;
      }
    } else {
      if ((gor - dol >= 2) && ((lstg > iles) || (lstd > ilew))) {
        printf("%lld\n", (pra - lew + 1) * (gor - dol));
        continue;
      }
    }
    if ((pra - lew >= 2) && ((lstl > iled) || (lstp > ilea))) {
      printf("%lld\n", (pra - lew) * (gor - dol + 1));
      continue;
    }
    if ((gor - dol >= 2) && ((lstg > iles) || (lstd > ilew))) {
      printf("%lld\n", (pra - lew + 1) * (gor - dol));
      continue;
    }
    printf("%lld\n", (pra - lew + 1) * (gor - dol + 1));
  }
  return 0;
}
