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 = 4e5 + 100;
struct node {
  long long minx, miny, maxx, maxy;
} aa[maxn], bb[maxn];
char a[maxn];
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    cin >> (a + 1);
    a[0] = ' ';
    int len = strlen(a) - 1;
    long long x = 0, y = 0;
    for (int i = 1; i <= len; i++) {
      if (a[i] == 'D') {
        x++;
        a[i] = 'A';
      } else if (a[i] == 'S') {
        y--;
        a[i] = 'W';
      } else if (a[i] == 'A') {
        x--;
        a[i] = 'D';
      } else if (a[i] == 'W') {
        a[i] = 'S';
        y++;
      }
      aa[i].maxx = max(aa[i - 1].maxx, x);
      aa[i].maxy = max(aa[i - 1].maxy, y);
      aa[i].minx = min(aa[i - 1].minx, x);
      aa[i].miny = min(aa[i - 1].miny, y);
    }
    bb[len].maxx = bb[len].minx = x;
    bb[len].maxy = bb[len].miny = y;
    for (int i = len; i >= 1; i--) {
      if (a[i] == 'D') {
        x++;
      }
      if (a[i] == 'S') {
        y--;
      }
      if (a[i] == 'A') {
        x--;
      }
      if (a[i] == 'W') {
        y++;
      }
      bb[i - 1].maxx = max(bb[i].maxx, x);
      bb[i - 1].maxy = max(bb[i].maxy, y);
      bb[i - 1].minx = min(bb[i].minx, x);
      bb[i - 1].miny = min(bb[i].miny, y);
    }
    long long ans = 999999999999, xx, yy, xxx, yyy;
    for (int i = 0; i < len; i++) {
      xx = min(aa[i].minx, bb[i].minx - 1);
      yy = min(aa[i].miny, bb[i].miny);
      xxx = max(aa[i].maxx, bb[i].maxx - 1);
      yyy = max(aa[i].maxy, bb[i].maxy);
      ans = min(ans, (xxx - xx + 1) * (yyy - yy + 1));
      xx = min(aa[i].minx, bb[i].minx + 1);
      yy = min(aa[i].miny, bb[i].miny);
      xxx = max(aa[i].maxx, bb[i].maxx + 1);
      yyy = max(aa[i].maxy, bb[i].maxy);
      ans = min(ans, (xxx - xx + 1) * (yyy - yy + 1));
      xx = min(aa[i].minx, bb[i].minx);
      yy = min(aa[i].miny, bb[i].miny - 1);
      xxx = max(aa[i].maxx, bb[i].maxx);
      yyy = max(aa[i].maxy, bb[i].maxy - 1);
      ans = min(ans, (xxx - xx + 1) * (yyy - yy + 1));
      xx = min(aa[i].minx, bb[i].minx);
      yy = min(aa[i].miny, bb[i].miny + 1);
      xxx = max(aa[i].maxx, bb[i].maxx);
      yyy = max(aa[i].maxy, bb[i].maxy + 1);
      ans = min(ans, (xxx - xx + 1) * (yyy - yy + 1));
    }
    cout << ans << endl;
  }
  return 0;
}
