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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt) {
    string s;
    cin >> s;
    int t, b, l, r;
    int t0, t1, b0, b1, l0, l1, r0, r1;
    t = b = l = r = t0 = t1 = b0 = b1 = l0 = l1 = r0 = r1 = 0;
    int x = 0, y = 0;
    for (int i = 0; i < s.size(); i++) {
      char c = s[i];
      switch (c) {
        case 'W':
          y += 1;
          if (y > t) {
            t = y;
            t0 = i;
            t1 = 0;
          } else if (y == t) {
            t1 = i;
          }
          break;
        case 'S':
          y -= 1;
          if (y < b) {
            b = y;
            b0 = i;
            b1 = 0;
          } else if (y == b) {
            b1 = i;
          }
          break;
        case 'A':
          x -= 1;
          if (x < l) {
            l = x;
            l0 = i;
            l1 = 0;
          } else if (x == l) {
            l1 = i;
          }
          break;
        case 'D':
          x += 1;
          if (x > r) {
            r = x;
            r0 = i;
            r1 = 0;
          } else if (x == r) {
            r1 = i;
          }
          break;
      }
    }
    long long sq = (t - b + 1) * (r - l + 1);
    if (t - b > 1 && ((!t1 && !b1) || (t1 && !b1 && (t1 < b0 || t0 > b0)) ||
                      (b1 && !t1 && (b1 < t0 || b0 > t0)) ||
                      (b1 && t1 && (b1 < t0 || t1 < b0)))) {
      sq = (long long)(t - b) * (r - l + 1);
    }
    if (r - l > 1 && ((!r1 && !l1) || (r1 && !l1 && (r1 < l0 || r0 > l0)) ||
                      (l1 && !r1 && (l1 < r0 || l0 > r0)) ||
                      (l1 && r1 && (l1 < r0 || r1 < l0)))) {
      sq = min(sq, (long long)(t - b + 1) * (r - l));
    }
    cout << sq << '\n';
    tt--;
  }
}
