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 long long INF = 0x3f3f3f3f3f3f3f3f;
long long T, len;
string s;
long long psa[200005];
long long psw[200005];
int main() {
  ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    long long dad = 0, dws = 0;
    long long mxa = 0, mia = 0, mxw = 0, miw = 0;
    long long axf = -1, axl = INF, aif = -1, ail = INF, wxf = -1, wxl = INF,
              wif = -1, wil = INF;
    cin >> s;
    len = s.length();
    s = ' ' + s;
    long long tx = 0, ty = 0;
    for (long long i = 1; i <= len; i++) {
      if (s[i] == 'W') ty--;
      if (s[i] == 'S') ty++;
      if (s[i] == 'A') tx--;
      if (s[i] == 'D') tx++;
      psa[i] = tx;
      psw[i] = ty;
      mxa = max(mxa, tx);
      mia = min(mia, tx);
      mxw = max(mxw, ty);
      miw = min(miw, ty);
    }
    for (long long i = 0; i <= len; i++) {
      if (psa[i] == mxa) axl = i;
      if (psa[i] == mia) ail = i;
      if (psw[i] == mxw) wxl = i;
      if (psw[i] == miw) wil = i;
    }
    for (long long i = len; i >= 0; i--) {
      if (psa[i] == mxa) axf = i;
      if (psa[i] == mia) aif = i;
      if (psw[i] == mxw) wxf = i;
      if (psw[i] == miw) wif = i;
    }
    if (axl < aif || ail < axf) dad = 1;
    if (wxl < wif || wil < wxf) dws = 1;
    mxa++;
    mxw++;
    if (mxa - mia <= 2) dad = 0;
    if (mxw - miw <= 2) dws = 0;
    cout << min((mxa - mia - dad) * (mxw - miw),
                (mxa - mia) * (mxw - miw - dws))
         << endl;
  }
  return 0;
}
