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 s[200010];
int minx[200010], maxx[200010], miny[200010], maxy[200010];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s", s);
    int n = strlen(s);
    int x = 0, y = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'W')
        y++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'A')
        x--;
      else if (s[i] == 'D')
        x++;
      minx[i] = maxx[i] = x, miny[i] = maxy[i] = y;
    }
    for (int i = n - 2; i >= 0; i--) {
      minx[i] = min(minx[i], minx[i + 1]);
      maxx[i] = max(maxx[i], maxx[i + 1]);
      miny[i] = min(miny[i], miny[i + 1]);
      maxy[i] = max(maxy[i], maxy[i + 1]);
    }
    long long delx = (max(maxx[0], 0) - min(minx[0], 0) + 1ll);
    long long dely = (max(maxy[0], 0) - min(miny[0], 0) + 1ll);
    long long ans = delx * dely;
    x = 0, y = 0;
    int max1 = 0, min1 = 0, max2 = 0, min2 = 0;
    for (int i = 0; i < n; i++) {
      int a = min(min1, minx[i] - 1), b = max(max1, maxx[i] - 1);
      a = min(a, x - 1), b = max(b, x - 1);
      ans = min(ans, (b - a + 1ll) * dely);
      a = min(min1, minx[i] + 1), b = max(max1, maxx[i] + 1);
      a = min(a, x + 1), b = max(b, x + 1);
      ans = min(ans, (b - a + 1ll) * dely);
      a = min(min2, miny[i] + 1), b = max(max2, maxy[i] + 1);
      a = min(a, y + 1), b = max(b, y + 1);
      ans = min(ans, delx * (b - a + 1ll));
      a = min(min2, miny[i] - 1), b = max(max2, maxy[i] - 1);
      a = min(a, y - 1), b = max(b, y - 1);
      ans = min(ans, delx * (b - a + 1ll));
      if (s[i] == 'W')
        y++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'A')
        x--;
      else if (s[i] == 'D')
        x++;
      min1 = min(min1, x), max1 = max(max1, x);
      min2 = min(min2, y), max2 = max(max2, y);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
