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 = 200010;
long long mw, ma, ms, md;
vector<int> cntw, cnta, cnts, cntd;
char s[MAXN];
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  for (int _ = (1); _ <= (T); _ += (1)) {
    mw = ma = ms = md = 0;
    cntw.clear();
    cnta.clear();
    cnts.clear();
    cntd.clear();
    cntw.push_back(0);
    cnta.push_back(0);
    cnts.push_back(0);
    cntd.push_back(0);
    cin >> s;
    int len = strlen(s);
    int x = 0, y = 0;
    for (int i = (0); i <= (len - 1); i += (1)) {
      if (s[i] == 'W') {
        y++;
        if (y > mw) {
          mw = y;
          cntw.clear();
          cntw.push_back(i);
        } else if (y == mw) {
          cntw.push_back(i);
        }
      } else if (s[i] == 'A') {
        x--;
        if (x < ma) {
          ma = x;
          cnta.clear();
          cnta.push_back(i);
        } else if (x == ma) {
          cnta.push_back(i);
        }
      } else if (s[i] == 'S') {
        y--;
        if (y < ms) {
          ms = y;
          cnts.clear();
          cnts.push_back(i);
        } else if (y == ms) {
          cnts.push_back(i);
        }
      } else if (s[i] == 'D') {
        x++;
        if (x > md) {
          md = x;
          cntd.clear();
          cntd.push_back(i);
        } else if (x == md) {
          cntd.push_back(i);
        }
      }
    }
    long long res = (mw - ms + 1) * (md - ma + 1);
    if (mw - ms > 1 && cntw.size() > 0 && cnts.size() > 0 &&
        (cntw[cntw.size() - 1] < cnts[0] || cnts[cnts.size() - 1] < cntw[0]))
      res = min(res, (mw - ms) * (md - ma + 1));
    if (md - ma > 1 && cnta.size() > 0 && cntd.size() > 0 &&
        (cnta[cnta.size() - 1] < cntd[0] || cntd[cntd.size() - 1] < cnta[0]))
      res = min(res, (mw - ms + 1) * (md - ma));
    cout << res << endl;
  }
  return 0;
}
