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 int mod = 1e9 + 7;
const long long int N = 1e6 + 5;
long long int t, n, m, a[N] = {0}, prefx[N], prefy[N];
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, x, y;
  cin >> t;
  while (t--) {
    cin >> s;
    prefy[0] = prefx[0] = 0;
    long long int mxx = 0, mxy = 0, mnx = 0, mny = 0;
    for (i = 0; i < ((long long int)s.size()); i++) {
      if (i > 0) {
        prefx[i] = prefx[i - 1];
        prefy[i] = prefy[i - 1];
      }
      if (s[i] == 'W') prefy[i]++;
      if (s[i] == 'S') prefy[i]--;
      if (s[i] == 'A') prefx[i]--;
      if (s[i] == 'D') prefx[i]++;
      mxx = max(mxx, prefx[i]);
      mxy = max(mxy, prefy[i]);
      mnx = min(mnx, prefx[i]);
      mny = min(mny, prefy[i]);
    }
    long long int dx = mxx - mnx + 1, dy = mxy - mny + 1;
    long long int ans = dx * dy;
    long long int lastmin = -1e10, lastmax = -1e10, firstmax = 1e10,
                  firstmin = 1e10;
    for (i = 0; i < ((long long int)s.size()); i++) {
      if (s[i] == 'W' || s[i] == 'S') continue;
      if (prefx[i] == mxx) {
        lastmax = max(lastmax, i);
        firstmax = min(firstmax, i);
      }
      if (prefx[i] == mnx) {
        lastmin = max(lastmin, i);
        firstmin = min(firstmin, i);
      }
    }
    if (mxx == 0) {
      lastmax = max(lastmax, -1ll);
      firstmax = min(firstmax, -1ll);
    }
    if (mnx == 0) {
      lastmin = max(lastmin, -1ll);
      firstmin = min(firstmin, -1ll);
    }
    if (dx > 2 && (lastmax < firstmin || firstmax > lastmin))
      ans = min(ans, dy * (dx - 1));
    lastmin = -1e10, lastmax = -1e10, firstmax = 1e10, firstmin = 1e10;
    for (i = 0; i < ((long long int)s.size()); i++) {
      if (s[i] == 'A' || s[i] == 'D') continue;
      if (prefy[i] == mxy) {
        lastmax = max(lastmax, i);
        firstmax = min(firstmax, i);
      }
      if (prefy[i] == mny) {
        lastmin = max(lastmin, i);
        firstmin = min(firstmin, i);
      }
    }
    if (mxy == 0) {
      lastmax = max(lastmax, -1ll);
      firstmax = min(firstmax, -1ll);
    }
    if (mny == 0) {
      lastmin = max(lastmin, -1ll);
      firstmin = min(firstmin, -1ll);
    }
    if (dy > 2 && (lastmax < firstmin || firstmax > lastmin))
      ans = min(ans, dx * (dy - 1));
    cout << ans << '\n';
  }
  return 0;
}
