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 = 200005;
int len, flag = 0;
string s;
long long solve(char c1, char c2) {
  int Min = 0, Max = 0, sum = 0;
  int p_lastMax = -1, p_firstMax = -1, p_lastMin = -1, p_firstMin = -1;
  int c = 0;
  for (int i = 0; i < len; i++) {
    bool ys = 0;
    if (s[i] == c1) sum--, ys = 1;
    if (s[i] == c2) sum++, ys = 1;
    if (ys) {
      c++;
      if (sum == Max) {
        p_lastMax = i;
      }
      if (sum > Max) {
        p_firstMax = p_lastMax = i;
        Max = sum;
      }
      if (sum == Min) {
        p_lastMin = i;
      }
      if (sum < Min) {
        p_firstMin = p_lastMin = i;
        Min = sum;
      }
    }
  }
  long long tmp = Max - Min + 1;
  if ((c > 1) && (p_lastMax < p_firstMin || p_firstMax > p_lastMin)) {
    tmp--;
    flag++;
  }
  return tmp;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    flag = 0;
    cin >> s;
    len = s.size();
    long long W = solve('W', 'S');
    long long H = solve('A', 'D');
    long long ans;
    if (flag == 2)
      ans = min(W * (H + 1), (W + 1) * H);
    else
      ans = W * H;
    cout << ans << endl;
  }
  return 0;
}
