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 = 200005;
const char cc[4] = {'W', 'S', 'A', 'D'};
const int xx[4] = {-1, 1, 0, 0};
const int yy[4] = {0, 0, -1, 1};
int n;
char aa[N], a[N];
int px[N], py[N];
int minpx[N], maxpx[N], minpy[N], maxpy[N];
int sx[N], sy[N];
int minsx[N], maxsx[N], minsy[N], maxsy[N];
void solv() {
  cin >> aa;
  n = strlen(aa);
  for (int i = 1; i <= n; ++i) a[i] = aa[i - 1];
  for (int i = 1; i <= n; ++i) {
    px[i] = px[i - 1];
    py[i] = py[i - 1];
    for (int j = 0; j < 4; ++j) {
      if (a[i] == cc[j]) {
        px[i] += xx[j];
        py[i] += yy[j];
      }
    }
    maxpx[i] = max(maxpx[i - 1], px[i]);
    minpx[i] = min(minpx[i - 1], px[i]);
    maxpy[i] = max(maxpy[i - 1], py[i]);
    minpy[i] = min(minpy[i - 1], py[i]);
  }
  sx[n + 1] = 0;
  sy[n + 1] = 0;
  maxsx[n + 1] = maxsy[n + 1] = minsx[n + 1] = minsy[n + 1] = 0;
  int maxx = 0, minx = 0, maxy = 0, miny = 0;
  for (int i = n; i >= 1; --i) {
    sx[i] = sx[i + 1];
    sy[i] = sy[i + 1];
    for (int j = 0; j < 4; ++j) {
      if (a[i] == cc[j]) {
        sx[i] += xx[j];
        sy[i] += yy[j];
      }
    }
    maxx = max(maxx, sx[i]);
    minx = min(minx, sx[i]);
    maxy = max(maxy, sy[i]);
    miny = min(miny, sy[i]);
    maxsx[i] = sx[i] - minx;
    minsx[i] = sx[i] - maxx;
    maxsy[i] = sy[i] - miny;
    minsy[i] = sy[i] - maxy;
  }
  long long ans = (maxpx[n] - minpx[n] + 1) * 1LL * (maxpy[n] - minpy[n] + 1);
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j < 4; ++j) {
      int maxx = maxpx[i];
      int minx = minpx[i];
      int maxy = maxpy[i];
      int miny = minpy[i];
      maxx = max(maxx, px[i] + xx[j]);
      minx = min(minx, px[i] + xx[j]);
      maxy = max(maxy, py[i] + yy[j]);
      miny = min(miny, py[i] + yy[j]);
      maxx = max(maxx, px[i] + xx[j] + maxsx[i + 1]);
      minx = min(minx, px[i] + xx[j] + minsx[i + 1]);
      maxy = max(maxy, py[i] + yy[j] + maxsy[i + 1]);
      miny = min(miny, py[i] + yy[j] + minsy[i + 1]);
      ans = min(ans, (maxx - minx + 1) * 1LL * (maxy - miny + 1));
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  int tt;
  cin >> tt;
  while (tt--) solv();
  return 0;
}
