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>
#pragma GCC optimize("O2")
using namespace std;
const int MAX = 1e5 + 5;
const long long MAX2 = 11;
const long long MOD = 1000000007;
const long long INF = 1e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 450;
long long tc, x, y, w, a, s, d, wse, ad;
string mv;
bool ss, aa, dd, ww;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> mv;
    x = y = w = a = s = d = 0;
    ww = ss = aa = dd = 0;
    for (auto i : mv) {
      if (i == 'W') {
        ++y;
        if (y > w) {
          w = y;
          ww = 1, ss = 0;
        } else if (y == w)
          ww = 1;
      } else if (i == 'S') {
        --y;
        if (y < s) {
          s = y;
          ss = 1, ww = 0;
        } else if (y == s)
          ss = 1;
      } else if (i == 'D') {
        ++x;
        if (x > d) {
          d = x;
          dd = 1, aa = 0;
        } else if (x == d)
          dd = 1;
      } else if (i == 'A') {
        --x;
        if (x < a) {
          a = x;
          aa = 1, dd = 0;
        } else if (x == a)
          aa = 1;
      } else
        return 1;
    }
    wse = w - s + 1;
    ad = d - a + 1;
    if (wse < ad) {
      if (ww && ss) {
        if (!(aa && dd) && ad > 2) ad--;
      } else if (wse > 2)
        wse--;
      else if (!(aa && dd) && ad > 2)
        ad--;
    } else {
      if (aa && dd) {
        if (!(ww && ss) && wse > 2) wse--;
      } else if (ad > 2)
        ad--;
      else if (!(ww && ss) && wse > 2)
        wse--;
    }
    cout << wse * ad << "\n";
  }
  return 0;
}
