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;
inline void Init() {}
int main() {
  Init();
  ios::sync_with_stdio(false);
  cin.tie(0);
  int x;
  cin >> x;
  while (x--) {
    string a;
    cin >> a;
    long long ans = 0x3f3f3f3f;
    int xma = 0, yma = 0, ymi = 0, xmi = 0, x = 0, y = 0;
    int xmai1 = 0, ymai1 = 0, ymii1 = 0, xmii1 = 0;
    int xmai2 = 0, ymai2 = 0, ymii2 = 0, xmii2 = 0;
    for (int i = 0; i < a.length(); i++) {
      if (a[i] == 'W')
        y++;
      else if (a[i] == 'S')
        y--;
      else if (a[i] == 'A')
        x--;
      else if (a[i] == 'D')
        x++;
      if (x < xmi) {
        xmi = min(x, xmi);
        xmii2 = xmii1 = i;
      }
      if (x > xma) {
        xma = max(x, xma);
        xmai2 = xmai1 = i;
      }
      if (x == xmi) {
        xmii2 = i;
      }
      if (x == xma) {
        xmai2 = i;
      }
      if (y < ymi) {
        ymi = min(y, ymi);
        ymii2 = ymii1 = i;
      }
      if (y > yma) {
        yma = max(y, yma);
        ymai2 = ymai1 = i;
      }
      if (y == ymi) {
        ymii2 = i;
      }
      if (y == yma) {
        ymai2 = i;
      }
    }
    bool flag = false, flag1 = false;
    if ((xmai2 < xmii1 || xmai1 > xmii2) && xma - xmi > 1) flag = 1;
    if ((ymai2 < ymii1 || ymai1 > ymii2) && yma - ymi > 1) flag1 = 1;
    long long n = (xma - xmi);
    long long m = (yma - ymi);
    ans = (n + 1) * (m + 1);
    if (flag && flag1)
      ans = min(n * (m + 1), (n + 1) * m);
    else if (flag)
      ans = (n) * (m + 1);
    else if (flag1)
      ans = (n + 1) * (m);
    cout << ans << endl;
  }
}
