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 maxn = 2e5 + 5;
string cmds;
struct Pos {
  long long x, y;
  Pos() {}
  Pos(long long x, long long y) : x(x), y(y) {}
} lp[maxn], rp[maxn], p[maxn];
long long mnx[maxn], mxx[maxn], mny[maxn], mxy[maxn];
void calc() {
  int n = cmds.size();
  long long nx = 1, ny = 1;
  lp[0] = rp[0] = p[0] = Pos(1, 1);
  for (int i = 1; i <= cmds.size(); i++) {
    lp[i] = lp[i - 1], rp[i] = rp[i - 1];
    if (cmds[i - 1] == 'W') lp[i].x = min(lp[i].x, --nx);
    if (cmds[i - 1] == 'S') rp[i].x = max(rp[i].x, ++nx);
    if (cmds[i - 1] == 'A') lp[i].y = min(lp[i].y, --ny);
    if (cmds[i - 1] == 'D') rp[i].y = max(rp[i].y, ++ny);
    p[i] = Pos(nx, ny);
  }
  mnx[n] = mxx[n] = p[n].x;
  mny[n] = mxy[n] = p[n].y;
  for (int i = n - 1; i >= 0; i--)
    mnx[i] = min(mnx[i + 1], p[i].x), mxx[i] = max(mxx[i + 1], p[i].x),
    mny[i] = min(mny[i + 1], p[i].y), mxy[i] = max(mxy[i + 1], p[i].y);
  long long ans = (rp[n].x - lp[n].x + 1) * (rp[n].y - lp[n].y + 1);
  for (int i = 0; i < n; i++) {
    Pos pos = p[i];
    long long x1 = min(lp[i].x, mnx[i + 1]), y1 = min(lp[i].y, mny[i + 1]),
              x2 = max(rp[i].x, mxx[i + 1]), y2 = max(rp[i].y, mxy[i + 1]);
    if (cmds[i] == 'W') {
      pos.x++;
      x2 = max(pos.x, max(rp[i].x, mxx[i + 1] + 1));
      x1 = min(lp[i].x, mnx[i + 1] + 1);
    }
    if (cmds[i] == 'S') {
      pos.x--;
      x1 = min(min(lp[i].x, mnx[i + 1] - 1), pos.x);
      x2 = max(rp[i].x, mxx[i + 1] - 1);
    }
    if (cmds[i] == 'A') {
      pos.y++;
      y2 = max(max(rp[i].y, mxy[i + 1] + 1), pos.y);
      y1 = min(lp[i].y, mny[i + 1] + 1);
    }
    if (cmds[i] == 'D') {
      pos.y--;
      y1 = min(min(lp[i].y, mny[i + 1] - 1), pos.y);
      y2 = max(rp[i].y, mxy[i + 1] - 1);
    }
    ans = min(ans, (x2 - x1 + 1) * (y2 - y1 + 1));
  }
  printf("%I64d\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    cin >> cmds;
    calc();
  }
  return 0;
}
