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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int yatay, dikey;
    long long k = 0;
    long long minny = 0;
    long long maxxy = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'D')
        k++;
      else if (s[i] == 'A')
        k--;
      minny = min(minny, k);
      maxxy = max(maxxy, k);
    }
    int num = 0;
    int last = 0;
    k = 0;
    if (k == minny && last != -1) {
      num++;
      last = -1;
    }
    if (k == maxxy && last != 1) {
      num++;
      last = 1;
    }
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'D')
        k++;
      else if (s[i] == 'A')
        k--;
      if (k == minny && last != -1) {
        num++;
        last = -1;
      }
      if (k == maxxy && last != 1) {
        num++;
        last = 1;
      }
    }
    yatay = num < 3;
    if (maxxy <= minny + 1) yatay = 0;
    long long minnd = 0;
    long long maxxd = 0;
    k = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'W')
        k++;
      else if (s[i] == 'S')
        k--;
      minnd = min(minnd, k);
      maxxd = max(maxxd, k);
    }
    num = 0;
    last = 0;
    k = 0;
    if (k == minnd && last != -1) {
      num++;
      last = -1;
    }
    if (k == maxxd && last != 1) {
      num++;
      last = 1;
    }
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'W')
        k++;
      else if (s[i] == 'S')
        k--;
      if (k == minnd && last != -1) {
        num++;
        last = -1;
      }
      if (k == maxxd && last != 1) {
        num++;
        last = 1;
      }
    }
    dikey = (num < 3);
    if (maxxd <= minnd + 1) dikey = 0;
    cout << min((maxxy - minny + 1) * (maxxd - minnd + 1) -
                    yatay * ((maxxd - minnd + 1)),
                (maxxy - minny + 1) * (maxxd - minnd + 1) -
                    dikey * (maxxy - minny + 1))
         << endl;
  }
}
