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>
char s[210000];
int T, n, xa, xi, ya, yi, xl, xr, yl, yr, x[210000], y[210000], fxi[210000],
    fxa[210000], fyi[210000], fya[210000], gxi[210000], gxa[210000],
    gyi[210000], gya[210000];
long long ans, te;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (int i = 1; i <= n; i++) {
      x[i] = x[i - 1];
      y[i] = y[i - 1];
      if (s[i] == 'A') y[i]--;
      if (s[i] == 'D') y[i]++;
      if (s[i] == 'W') x[i]--;
      if (s[i] == 'S') x[i]++;
    }
    xa = xi = ya = yi = xl = xr = yl = yr = 0;
    for (int i = 1; i <= n; i++) {
      if (x[i] >= xa) {
        xa = x[i];
        xl = i;
      }
      if (y[i] >= ya) {
        ya = y[i];
        yl = i;
      }
      if (x[i] < xi) {
        xi = x[i];
        xr = i;
      }
      if (y[i] < yi) {
        yi = y[i];
        yr = i;
      }
    }
    ans = (xa - xi + 1) * 1ll * (ya - yi + 1);
    if (xl < xr && xi + 1 < xa) {
      te = (xa - xi) * 1ll * (ya - yi + 1);
      if (te < ans) ans = te;
    }
    if (yl < yr && yi + 1 < ya) {
      te = (xa - xi + 1) * 1ll * (ya - yi);
      if (te < ans) ans = te;
    }
    for (int i = 1; i <= n; i++) {
      x[i] *= -1;
      y[i] *= -1;
    }
    xa = xi = ya = yi = xl = xr = yl = yr = 0;
    for (int i = 1; i <= n; i++) {
      if (x[i] >= xa) {
        xa = x[i];
        xl = i;
      }
      if (y[i] >= ya) {
        ya = y[i];
        yl = i;
      }
      if (x[i] < xi) {
        xi = x[i];
        xr = i;
      }
      if (y[i] < yi) {
        yi = y[i];
        yr = i;
      }
    }
    if (xl < xr && xi + 1 < xa) {
      te = (xa - xi) * 1ll * (ya - yi + 1);
      if (te < ans) ans = te;
    }
    if (yl < yr && yi + 1 < ya) {
      te = (xa - xi + 1) * 1ll * (ya - yi);
      if (te < ans) ans = te;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
