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;
char s[200005], t[200005];
int n, w, h, u, d, l, r, x, y, pu, pd, pl, pr;
void insert(int pos, char ch) {
  int len = strlen(t + 1);
  for (int i = len; i >= pos; i--) t[i + 1] = t[i];
  t[pos] = ch;
}
void calc() {
  int len = strlen(t + 1), u = 0, d = 0, l = 0, r = 0, x = 0, y = 0;
  for (int i = 1; i <= len; i++) {
    if (t[i] == 'W') y++;
    if (t[i] == 'S') y--;
    if (t[i] == 'A') x--;
    if (t[i] == 'D') x++;
    if (y > u) u = y;
    if (y < d) d = y;
    if (x > r) r = x;
    if (x < l) l = x;
  }
  w = r - l + 1;
  h = u - d + 1;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    u = 0, d = 0, l = 0, r = 0, x = 0, y = 0;
    pu = pd = pl = pr = 0;
    memset(s, 0, sizeof s);
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'W') y++;
      if (s[i] == 'S') y--;
      if (s[i] == 'A') x--;
      if (s[i] == 'D') x++;
      if (y > u) u = y, pu = i;
      if (y < d) d = y, pd = i;
      if (x > r) r = x, pr = i;
      if (x < l) l = x, pl = i;
    }
    w = r - l + 1;
    h = u - d + 1;
    long long ans = (long long)w * h;
    memset(t, 0, sizeof t);
    memcpy(t, s, sizeof s);
    if (pu) insert(pu, 'S');
    calc();
    ans = min(ans, 1ll * w * h);
    memset(t, 0, sizeof t);
    memcpy(t, s, sizeof s);
    if (pd) insert(pd, 'W');
    calc();
    ans = min(ans, 1ll * w * h);
    memset(t, 0, sizeof t);
    memcpy(t, s, sizeof s);
    if (pr) insert(pr, 'A');
    calc();
    ans = min(ans, 1ll * w * h);
    memset(t, 0, sizeof t);
    memcpy(t, s, sizeof s);
    if (pl) insert(pl, 'D');
    calc();
    ans = min(ans, 1ll * w * h);
    cout << ans << '\n';
  }
}
