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;
const int N = 2e5 + 10;
char s[N];
int pre[N], h[N];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s + 1);
    memset(pre, 0, sizeof(pre));
    memset(h, 0, sizeof(h));
    int x, y, l, r, u, d;
    x = y = l = r = u = d = 0;
    int len = strlen(s + 1);
    for (int i = 1; i <= len; i++) {
      if (s[i] == 'W')
        y++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'A')
        x--;
      else
        x++;
      l = min(l, x), r = max(r, x), u = max(u, y), d = min(d, y);
    }
    long long ans = 1ll * (r - l + 1) * (u - d + 1);
    string a, b;
    for (int i = 1; i <= len; i++)
      if (s[i] == 'W' || s[i] == 'S')
        a += s[i];
      else
        b += s[i];
    int la = a.length(), lb = b.length();
    for (int i = 1; i <= la; i++)
      if (a[i - 1] == 'W')
        pre[i] = pre[i - 1] + 1;
      else
        pre[i] = pre[i - 1] - 1;
    for (int i = 1; i <= lb; i++)
      if (b[i - 1] == 'D')
        h[i] = h[i - 1] + 1;
      else
        h[i] = h[i - 1] - 1;
    int pl1 = 0, pl2 = 0, pr1 = 0, pr2 = 0;
    int mi = 0, ma = 0;
    for (int i = 1; i <= la; i++)
      if (pre[i] > ma) {
        ma = pre[i];
        pl1 = i;
      }
    for (int i = la; i >= 1; i--)
      if (pre[i] == ma) {
        pr1 = i;
        break;
      }
    for (int i = 1; i <= la; i++)
      if (pre[i] < mi) {
        mi = pre[i];
        pl2 = i;
      }
    for (int i = la; i >= 1; i--)
      if (pre[i] == mi) {
        pr2 = i;
        break;
      }
    if (pr2 < pl1 && la > 1 || pl2 > pr1 && la > 1)
      ans = min(ans, (u - d) * 1ll * (r - l + 1));
    pl1 = pl2 = pr1 = pr2 = 0;
    mi = 0, ma = 0;
    for (int i = 1; i <= lb; i++)
      if (h[i] > ma) {
        ma = h[i];
        pl1 = i;
      }
    for (int i = lb; i >= 1; i--)
      if (h[i] == ma) {
        pr1 = i;
        break;
      }
    for (int i = 1; i <= lb; i++)
      if (h[i] < mi) {
        mi = h[i];
        pl2 = i;
      }
    for (int i = lb; i >= 1; i--)
      if (h[i] == mi) {
        pr2 = i;
        break;
      }
    if (pr2 < pl1 && lb > 1 || pl2 > pr1 && lb > 1)
      ans = min(ans, (u - d + 1) * 1ll * (r - l));
    printf("%lld\n", ans);
  }
  return 0;
}
