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 x, y, l, r, u, d;
int lowl, highl, lowr, highr;
int lowu, highu, lowd, highd;
int nd, nu, nl, nr;
long long res, s;
string rb;
void Solve() {
  x = y = 0;
  l = r = u = d = 0;
  lowl = lowu = lowr = lowd = 0;
  highl = highu = highr = highd = 0;
  for (int i = 0; i < (int)rb.size(); i++) {
    if (rb[i] == 'S') {
      x++;
      if (x > d) {
        d = x;
        lowd = highd = i;
      } else if (x == d)
        highd = i;
    } else if (rb[i] == 'W') {
      x--;
      if (x < u) {
        u = x;
        lowu = highu = i;
      } else if (x == u)
        highu = i;
    } else if (rb[i] == 'A') {
      y--;
      if (y < l) {
        l = y;
        lowl = highl = i;
      } else if (y == l)
        highl = i;
    } else {
      y++;
      if (y > r) {
        r = y;
        lowr = highr = i;
      } else if (y == r)
        highr = i;
    }
  }
  res = (d - u + 1) * ((long long)r - l + 1);
  nd = d;
  nu = u;
  nd--;
  if (lowd <= highu) nu--;
  nu = min(nu, nd - 1);
  s = (nd - nu + 1) * ((long long)r - l + 1);
  res = min(res, s);
  nd = d;
  nu = u;
  nu++;
  if (lowu <= highd) nd++;
  nd = max(nd, nu + 1);
  s = (nd - nu + 1) * ((long long)r - l + 1);
  res = min(res, s);
  nr = r;
  nl = l;
  nr--;
  if (lowr <= highl) nl--;
  nl = min(nr - 1, nl);
  s = (d - u + 1) * ((long long)nr - nl + 1);
  res = min(res, s);
  nl = l;
  nr = r;
  nl++;
  if (lowl <= highr) nr++;
  nr = max(nl + 1, nr);
  s = (d - u + 1) * ((long long)nr - nl + 1);
  res = min(res, s);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    cin >> rb;
    Solve();
    cout << res << '\n';
  }
}
