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 + 7;
char s[N];
int n, m, i, j, k, T_T, x, y, nx, ny, mnx, mny, mxx, mxy, dx[4] = {1, -1, 0, 0},
                                                          dy[4] = {0, 0, 1, -1};
int Lx[N], Ly[N], Rx[N], Ry[N], Lmnx[N], Lmny[N], Lmxx[N], Lmxy[N], Rmnx[N],
    Rmny[N], Rmxx[N], Rmxy[N];
long long ans;
void update() {
  mnx = min(mnx, x);
  mxx = max(mxx, x);
  mny = min(mny, y);
  mxy = max(mxy, y);
}
int main() {
  for (scanf("%d", &T_T); T_T--;) {
    scanf("%s", s + 1), n = strlen(s + 1);
    mnx = mny = mxx = mxy = x = y = 0;
    for (i = 1; i <= n; ++i) {
      if (s[i] == 'W') x--;
      if (s[i] == 'S') x++;
      if (s[i] == 'A') y--;
      if (s[i] == 'D') y++;
      update();
      Lx[i] = x;
      Ly[i] = y;
      Lmnx[i] = mnx;
      Lmxx[i] = mxx;
      Lmny[i] = mny;
      Lmxy[i] = mxy;
    }
    ans = 1LL * (mxx - mnx + 1) * (mxy - mny + 1);
    mnx = mny = mxx = mxy = x = y = 0;
    for (i = n; i > 0; --i) {
      if (s[i] == 'S') x--;
      if (s[i] == 'W') x++;
      if (s[i] == 'D') y--;
      if (s[i] == 'A') y++;
      Rx[i] = x;
      Ry[i] = y;
      update();
      Rmnx[i] = mnx;
      Rmxx[i] = mxx;
      Rmny[i] = mny;
      Rmxy[i] = mxy;
    }
    for (i = 1; i < n; ++i) {
      nx = Lx[i];
      ny = Ly[i];
      for (k = 0; k < 4; ++k) {
        x = nx + dx[k];
        y = ny + dy[k];
        mnx = Lmnx[i];
        mny = Lmny[i];
        mxx = Lmxx[i];
        mxy = Lmxy[i];
        update();
        if (i == 5 && k == 0) {
        }
        mnx -= x;
        mxx -= x;
        mny -= y;
        mxy -= y;
        mxx = max(mxx, Rmxx[i + 1] - Rx[i + 1]);
        mnx = min(mnx, Rmnx[i + 1] - Rx[i + 1]);
        mxy = max(mxy, Rmxy[i + 1] - Ry[i + 1]);
        mny = min(mny, Rmny[i + 1] - Ry[i + 1]);
        if (i == 5 && k == 0) {
        }
        ans = min(ans, 1LL * (mxx - mnx + 1) * (mxy - mny + 1));
        if (1LL * (mxx - mnx + 1) * (mxy - mny + 1) == 6) {
        }
      }
    }
    printf("%lld\n", ans);
  }
}
