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;
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long maxu = 0;
    long long maxd = 0;
    long long maxl = 0;
    long long maxr = 0;
    long long iu = 0;
    long long id = 0;
    long long il = 0;
    long long ir = 0;
    long long ju = 0;
    long long jd = 0;
    long long jl = 0;
    long long jr = 0;
    long long v = 0;
    long long h = 0;
    for (long long i = 0; i < s.length(); i++) {
      if (s[i] == 'W') {
        v++;
        if (v > maxu) {
          iu = i;
          ju = i;
        }
        if (v == maxu) {
          ju = i;
        }
        maxu = max(maxu, v);
      } else if (s[i] == 'A') {
        h--;
        if (h < maxl) {
          il = i;
          jl = i;
        }
        if (h == maxl) {
          jl = i;
        }
        maxl = min(maxl, h);
      } else if (s[i] == 'S') {
        v--;
        if (v < maxd) {
          id = i;
          jd = i;
        }
        if (v == maxd) {
          jd = i;
        }
        maxd = min(maxd, v);
      } else if (s[i] == 'D') {
        h++;
        if (h > maxr) {
          ir = i;
          jr = i;
        }
        if (h == maxr) {
          jr = i;
        }
        maxr = max(maxr, h);
      }
    }
    long long i = maxu - maxd + 1;
    long long j = maxr - maxl + 1;
    long long ans = i * j;
    if (maxu - maxd > 1 && (jd < iu || ju < id)) {
      i--;
      ans = min(i * j, ans);
      i++;
    }
    if (maxr - maxl > 1 && (jl < ir || jr < il)) {
      j--;
      ans = min(i * j, ans);
      j++;
    }
    cout << ans << "\n";
  }
  return 0;
}
