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 = 2e5 + 50;
int t, len, x, y, dx[4] = {-1, 0, 0, 1}, dy[4] = {0, -1, 1, 0}, a[N], mxx, mnx,
                  mxy, mny;
char c[N];
long long ans;
void cmax(int &x, int y) { y > x ? x = y : 0; }
void cmin(int &x, int y) { y < x ? x = y : 0; }
bool check1() {
  x = 0;
  y = 0;
  int flag = 0;
  if (x == mxx) flag = 1;
  if (x == mnx) flag = 2;
  for (int i = 1; i <= len; i++) {
    x += dx[a[i]];
    y += dy[a[i]];
    if (x == mxx) {
      if (!flag)
        flag = 1;
      else if (flag == 2)
        flag = 3;
      else if (flag == 4)
        return 0;
    }
    if (x == mnx) {
      if (!flag)
        flag = 2;
      else if (flag == 1)
        flag = 4;
      else if (flag == 3)
        return 0;
    }
  }
  return 1;
}
bool check2() {
  x = 0;
  y = 0;
  int flag = 0;
  if (y == mxy) flag = 1;
  if (y == mny) flag = 2;
  for (int i = 1; i <= len; i++) {
    x += dx[a[i]];
    y += dy[a[i]];
    if (y == mxy) {
      if (!flag)
        flag = 1;
      else if (flag == 2)
        flag = 3;
      else if (flag == 4)
        return 0;
    }
    if (y == mny) {
      if (!flag)
        flag = 2;
      else if (flag == 1)
        flag = 4;
      else if (flag == 3)
        return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%s", c + 1);
    len = strlen(c + 1);
    x = 0;
    y = 0;
    for (int i = 1; i <= len; i++) {
      if (c[i] == 'W') a[i] = 0;
      if (c[i] == 'A') a[i] = 1;
      if (c[i] == 'D') a[i] = 2;
      if (c[i] == 'S') a[i] = 3;
    }
    mxx = mnx = mxy = mny = 0;
    for (int i = 1; i <= len; i++) {
      x += dx[a[i]];
      y += dy[a[i]];
      cmax(mxx, x);
      cmin(mnx, x);
      cmax(mxy, y);
      cmin(mny, y);
    }
    ans = 1ll * (mxx - mnx + 1) * (mxy - mny + 1);
    if (mxx - mnx + 1 >= 3 && check1())
      ans = min(ans, 1ll * (mxx - mnx) * (mxy - mny + 1));
    if (mxy - mny + 1 >= 3 && check2())
      ans = min(ans, 1ll * (mxx - mnx + 1) * (mxy - mny));
    cout << ans << endl;
  }
  return 0;
}
