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 maxn = 2e5 + 1000;
char s[maxn];
long long get(int f, long long up) {
  int ls = strlen(s);
  long long x = 0, y = 0;
  long long maxx = 0, maxy = 0, minx = 0, miny = 0;
  for (int i = 0; i < ls; i++) {
    if (s[i] == 'W')
      y++;
    else if (s[i] == 'S')
      y--;
    else if (s[i] == 'A')
      x--;
    else
      x++;
    if (f == 1) {
      if (up == x + 1 && s[i + 1] == 'D') {
        x--;
        f = 0;
      }
    } else if (f == 2) {
      if (up == x - 1 && s[i + 1] == 'A') {
        x++;
        f = 0;
      }
    } else if (f == 3) {
      if (up == y + 1 && s[i + 1] == 'W') {
        f = 0;
        y--;
      }
    } else if (f == 4) {
      if (up == y - 1 && s[i + 1] == 'S') {
        y++;
        f = 0;
      }
    }
    maxx = max(maxx, x);
    maxy = max(maxy, y);
    minx = min(minx, x);
    miny = min(miny, y);
  }
  return 1ll * (maxx + 1 - minx) * (maxy + 1 - miny);
}
int main() {
  int kase;
  scanf("%d", &kase);
  while (kase--) {
    scanf("%s", s);
    int x = 0, y = 0;
    int maxx = 0, maxy = 0, minx = 0, miny = 0;
    int ls = strlen(s);
    for (int i = 0; i < ls; i++) {
      if (s[i] == 'W')
        y++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'A')
        x--;
      else
        x++;
      maxx = max(maxx, x);
      maxy = max(maxy, y);
      minx = min(minx, x);
      miny = min(miny, y);
    }
    long long ans = 0;
    ans = 1ll * (maxx - minx + 1) * (maxy - miny + 1);
    ans = min(ans, get(1, maxx));
    ans = min(ans, get(2, minx));
    ans = min(ans, get(3, maxy));
    ans = min(ans, get(4, miny));
    cout << ans << endl;
  }
  return 0;
}
