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;
const int N = 2000010;
char tmpread[N];
string s;
void solve() {
  scanf("%s", tmpread);
  s = (string)tmpread;
  int mnx = 0, mxx = 0, mny = 0, mxy = 0;
  int x = 0, y = 0;
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == 'W')
      y++;
    else if (s[i] == 'S')
      y--;
    else if (s[i] == 'A')
      x--;
    else
      x++;
    mnx = min(mnx, x);
    mxx = max(mxx, x);
    mny = min(mny, y);
    mxy = max(mxy, y);
  }
  int a = 0, b = 0, c = 0, d = 0;
  x = y = 0;
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == 'W')
      y++;
    else if (s[i] == 'S')
      y--;
    else if (s[i] == 'A')
      x--;
    else
      x++;
    if (x == mnx) a++, x++;
    if (y == mny) c++, y++;
    if (x > mxx) a += 10;
    if (y > mxy) c += 10;
  }
  x = y = 0;
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == 'W')
      y++;
    else if (s[i] == 'S')
      y--;
    else if (s[i] == 'A')
      x--;
    else
      x++;
    if (x == mxx) b++, x--;
    if (y == mxy) d++, y--;
    if (x < mnx) b += 10;
    if (y < mny) d += 10;
  }
  if (mnx == 0) a += 10;
  if (mxx == 0) b += 10;
  if (mny == 0) c += 10;
  if (mxy == 0) d += 10;
  x = (mxx - mnx) + 1;
  y = (mxy - mny) + 1;
  bool canx = false;
  bool cany = false;
  if (x > 2 && (a < 2 || b < 2)) canx = true;
  if (y > 2 && (c < 2 || d < 2)) cany = true;
  long long ans = (long long)x * y;
  if (canx) ans = min(ans, (long long)(x - 1) * y);
  if (cany) ans = min(ans, (long long)x * (y - 1));
  printf("%lld\n", ans);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
