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() {
  long long int t, n, hs1, hs2, hp, vs1, vs2, vp, ths1, ths2, thp, tvs1, tvs2,
      tvp;
  cin >> t;
  string s;
  while (t-- > 0) {
    hs1 = 0;
    hs2 = 0;
    hp = 0;
    vs1 = 0;
    vs2 = 0;
    vp = 0;
    ths1 = 0;
    ths2 = 0;
    thp = 0;
    tvs1 = 0;
    tvs2 = 0;
    tvp = 0;
    cin >> s;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == 'W') {
        hp++;
        if (hp > hs1) hs1 = hp;
      }
      if (s[i] == 'S') {
        hp--;
        if (hp < hs2) hs2 = hp;
      }
      if (s[i] == 'A') {
        vp--;
        if (vp < vs2) vs2 = vp;
      }
      if (s[i] == 'D') {
        vp++;
        if (vp > vs1) vs1 = vp;
      }
    }
    bool fh1 = 0, fh2 = 0, fv1 = 0, fv2 = 0, h1 = 1, h2 = 1, v1 = 1, v2 = 1;
    if (thp == hs1) fh1 = 1;
    if (thp == hs2) fh2 = 1;
    if (tvp == vs2) fv2 = 1;
    if (tvp == vs1) fv1 = 1;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == 'W') {
        thp++;
        if (thp == hs1) fh1 = 1;
        if (fh2 == 1)
          if (thp == hs1) h1 = 0;
      }
      if (s[i] == 'S') {
        thp--;
        if (thp == hs2) fh2 = 1;
        if (fh1 == 1)
          if (thp == hs2) h2 = 0;
      }
      if (s[i] == 'A') {
        tvp--;
        if (tvp == vs2) fv2 = 1;
        if (fv1 == 1)
          if (tvp == vs2) v2 = 0;
      }
      if (s[i] == 'D') {
        tvp++;
        if (tvp == vs1) fv1 = 1;
        if (fv2 == 1)
          if (tvp == vs1) v1 = 0;
      }
    }
    long long int mi = (hs1 - hs2 + 1) * (vs1 - vs2 + 1);
    if (v1 + v2 == 1 && vs1 - vs2 != 1)
      mi = min(mi, (hs1 - hs2 + 1) * (vs1 - vs2));
    if (h1 + h2 == 1 && hs1 - hs2 != 1)
      mi = min(mi, (hs1 - hs2) * (vs1 - vs2 + 1));
    cout << mi << '\n';
  }
  return 0;
}
