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 int maxn = 7e5 + 7;
int p1 = 1, p2 = 1;
long long prex[maxn];
long long prey[maxn];
long long dpx[maxn][2];
long long dpy[maxn][2];
int main() {
  int T;
  cin >> T;
  while (T--) {
    p1 = p2 = 1;
    string s;
    cin >> s;
    int len = s.size();
    for (int i = 0; i <= len * 2; i++) {
      dpx[i][0] = dpx[i][1] = dpy[i][0] = dpy[i][1] = 0;
      prex[i] = prey[i] = 0;
    }
    for (int i = 0; i < len; i++) {
      if (s[i] == 'W') prey[p2] = 1, p2++;
      if (s[i] == 'S') prey[p2] = -1, p2++;
      if (s[i] == 'A') prex[p1] = -1, p1++;
      if (s[i] == 'D') prex[p1] = 1, p1++;
    }
    for (int i = p1; i >= 1; i--) {
      dpx[i][0] = min(prex[i], dpx[i + 1][0] + prex[i]);
      dpx[i][1] = max(prex[i], dpx[i + 1][1] + prex[i]);
    }
    for (int i = p2; i >= 1; i--) {
      dpy[i][0] = min(prey[i], dpy[i + 1][0] + prey[i]);
      dpy[i][1] = max(prey[i], dpy[i + 1][1] + prey[i]);
    }
    long long now = 0;
    long long lx = 0, rx = 0;
    long long lenx = 1e9;
    for (int i = 1; i <= p1; i++) {
      long long k1 = min(lx, now - prex[i]);
      long long k2 = max(rx, now - prex[i]);
      long long ll = min(k1, now + dpx[i + 1][0]);
      long long rr = max(k2, now + dpx[i + 1][1]);
      lenx = min((rr - ll + 1), lenx);
      k1 = min(lx, now + prex[i]);
      k2 = max(rx, now + prex[i]);
      ll = min(k1, now + dpx[i + 1][0]);
      rr = max(k2, now + dpx[i + 1][1]);
      lenx = min((rr - ll + 1), lenx);
      now += prex[i];
      lx = min(lx, now);
      rx = max(rx, now);
    }
    now = 0;
    long long ly = 0, ry = 0;
    long long leny = 1e9;
    for (int i = 1; i <= p2; i++) {
      long long k1 = min(ly, now - prey[i]);
      long long k2 = max(ry, now - prey[i]);
      long long ll = min(k1, now + dpy[i + 1][0]);
      long long rr = max(k2, now + dpy[i + 1][1]);
      leny = min((rr - ll + 1), leny);
      k1 = min(ly, now + prey[i]);
      k2 = max(ry, now + prey[i]);
      ll = min(k1, now + dpy[i + 1][0]);
      rr = max(k2, now + dpy[i + 1][1]);
      leny = min((rr - ll + 1), leny);
      now += prey[i];
      ly = min(ly, now);
      ry = max(ry, now);
    }
    long long ans = min(min((rx - lx + 1) * leny, (ry - ly + 1) * lenx),
                        (rx - lx + 1) * (ry - ly + 1));
    cout << ans << endl;
  }
  return 0;
}
