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() {
  int T;
  cin >> T;
  string s;
  int pSum1[200004];
  int pSum2[200004];
  int min1, min2, max1, max2;
  int fmin, fmax, lmin, lmax;
  int width, heigth, pwidth, pheigth;
  int i1, i2;
  long long ans;
  for (int i = 0; i < T; i++) {
    s.clear();
    cin >> s;
    pSum1[0] = 0;
    pSum2[0] = 0;
    min1 = 0;
    min2 = 0;
    max1 = 0;
    max2 = 0;
    i1 = 0, i2 = 0;
    for (int j = 0; j < s.length(); j++) {
      if (s[j] == 'W') {
        pSum1[i1 + 1] = pSum1[i1] + 1;
        if (pSum1[i1 + 1] > max1) max1 = pSum1[i1 + 1];
        i1++;
      }
      if (s[j] == 'S') {
        pSum1[i1 + 1] = pSum1[i1] - 1;
        if (pSum1[i1 + 1] < min1) min1 = pSum1[i1 + 1];
        i1++;
      }
      if (s[j] == 'A') {
        pSum2[i2 + 1] = pSum2[i2] - 1;
        if (pSum2[i2 + 1] < min2) min2 = pSum2[i2 + 1];
        i2++;
      }
      if (s[j] == 'D') {
        pSum2[i2 + 1] = pSum2[i2] + 1;
        if (pSum2[i2 + 1] > max2) max2 = pSum2[i2 + 1];
        i2++;
      }
    }
    width = max2 - min2 + 1;
    heigth = max1 - min1 + 1;
    pwidth = width;
    pheigth = heigth;
    fmin = -1;
    fmax = -1;
    for (int j = 0; j < i1 + 1; j++) {
      if (pSum1[j] == min1) {
        lmin = j;
        if (fmin == -1) fmin = j;
      }
      if (pSum1[j] == max1) {
        lmax = j;
        if (fmax == -1) fmax = j;
      }
    }
    if ((fmin > lmax || fmax > lmin) && heigth > 2) pheigth--;
    fmin = -1;
    fmax = -1;
    for (int j = 0; j < i2 + 1; j++) {
      if (pSum2[j] == min2) {
        lmin = j;
        if (fmin == -1) fmin = j;
      }
      if (pSum2[j] == max2) {
        lmax = j;
        if (fmax == -1) fmax = j;
      }
    }
    if ((fmin > lmax || fmax > lmin) && width > 2) pwidth--;
    ans = min((long long)heigth * pwidth, (long long)pheigth * width);
    cout << ans << '\n';
  }
}
