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 T;
char s[2000005];
int n;
int seq[2000005], top;
void Solve(int &nochange, int &onechange) {
  onechange = 1e9;
  int mmin = 0, mmax = 0, tmin = 0, tmax = 0;
  for (int i = 1; i <= top; i++)
    seq[i] += seq[i - 1], mmin = min(mmin, seq[i]), mmax = max(mmax, seq[i]);
  nochange = mmax - mmin + 1;
  int l = 0;
  if (mmin) {
    while (seq[l] != mmin) l++;
    for (int i = 1; i < l; i++) tmax = max(tmax, seq[i]);
    tmax = max(tmax, seq[l - 1] + 1);
    for (; l <= top; l++) tmax = max(tmax, seq[l] + 1);
    onechange = min(onechange, max(tmax, 0) - min(0, mmin + 1) + 1);
  }
  if (mmax) {
    l = 0;
    while (seq[l] != mmax) l++;
    for (int i = 1; i < l; i++) tmin = min(tmin, seq[i]);
    tmin = min(tmin, seq[l - 1] - 1);
    for (; l <= top; l++) tmin = min(tmin, seq[l] - 1);
    onechange = min(onechange, max(mmax - 1, 0) - min(0, tmin) + 1);
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    int a, b, c, d;
    scanf("%s", s + 1), n = strlen(s + 1);
    top = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'S')
        seq[++top] = -1;
      else if (s[i] == 'W')
        seq[++top] = 1;
    }
    Solve(a, b);
    top = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'A')
        seq[++top] = -1;
      else if (s[i] == 'D')
        seq[++top] = 1;
    }
    Solve(c, d);
    printf("%lld\n", min(1ll * a * c, min(1ll * b * c, 1ll * a * d)));
  }
  return 0;
}
