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 INF = 0x3f3f3f3f, N = 2e5 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int T, minx[N], maxx[N], miny[N], maxy[N];
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> T;
  while (T--) {
    cin >> s;
    int x = 0, y = 0;
    int n = s.size();
    for (int i = 0; i < n; i++) {
      if (s[i] == 'W')
        x--;
      else if (s[i] == 'S')
        x++;
      else if (s[i] == 'D')
        y++;
      else
        y--;
      minx[i + 1] = min(minx[i], x);
      maxx[i + 1] = max(maxx[i], x);
      miny[i + 1] = min(miny[i], y);
      maxy[i + 1] = max(maxy[i], y);
    }
    long long ans =
        (long long)(maxx[n] - minx[n] + 1) * (maxy[n] - miny[n] + 1);
    int nowmaxx = x, nowmaxy = y, nowminx = x, nowminy = y;
    for (int j = 0; j < 4; j++) {
      long long w = max(maxx[n] + dir[j][0], nowmaxx) -
                    min(minx[n] + dir[j][0], nowminx) + 1;
      long long h = max(maxy[n] + dir[j][1], nowmaxy) -
                    min(miny[n] + dir[j][1], nowminy) + 1;
      ans = min(ans, w * h);
    }
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == 'W')
        x++;
      else if (s[i] == 'S')
        x--;
      else if (s[i] == 'D')
        y--;
      else
        y++;
      nowmaxx = max(nowmaxx, x);
      nowminx = min(nowminx, x);
      nowmaxy = max(nowmaxy, y);
      nowminy = min(nowminy, y);
      for (int j = 0; j < 4; j++) {
        long long w = max(maxx[i] + dir[j][0], nowmaxx) -
                      min(minx[i] + dir[j][0], nowminx) + 1;
        long long h = max(maxy[i] + dir[j][1], nowmaxy) -
                      min(miny[i] + dir[j][1], nowminy) + 1;
        ans = min(ans, w * h);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
