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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long first = 0, second = 0;
    char c[200005];
    scanf("%s", c);
    long long Minx = 0, Maxx = 0, Miny = 0, Maxy = 0;
    vector<int> xr, yr;
    xr.push_back(0);
    yr.push_back(0);
    for (int i = 0; c[i] != 0; i++) {
      if (c[i] == 'W') first++;
      if (c[i] == 'S') first--;
      if (c[i] == 'A') second--;
      if (c[i] == 'D') second++;
      Minx = min(Minx, first);
      Miny = min(Miny, second);
      Maxx = max(Maxx, first);
      Maxy = max(Maxy, second);
      xr.push_back(first);
      yr.push_back(second);
    }
    int l = 1e9, r = xr.size() - 1;
    int okx = 0, oky = 0;
    for (int i = 0; i < xr.size(); i++) {
      if (xr[i] == Maxx) {
        l = min(l, i);
      }
    }
    if (l != 0 && xr[l - 1] != Minx) {
      int ook = 1;
      for (int i = l; i <= r; i++) {
        if (xr[i] == Minx) {
          ook = 0;
        }
      }
      if (ook) okx = 1;
    }
    l = 1e9, r = xr.size() - 1;
    for (int i = 0; i < xr.size(); i++) {
      if (xr[i] == Minx) {
        l = min(l, i);
      }
    }
    if (l != 0 && xr[l - 1] != Maxx) {
      int ook = 1;
      for (int i = l; i <= r; i++) {
        if (xr[i] == Maxx) {
          ook = 0;
        }
      }
      if (ook) okx = 1;
    }
    l = 1e9, r = yr.size() - 1;
    for (int i = 0; i < yr.size(); i++) {
      if (yr[i] == Miny) {
        l = min(l, i);
      }
    }
    if (l != 0 && yr[l - 1] != Maxy) {
      int ook = 1;
      for (int i = l; i <= r; i++) {
        if (yr[i] == Maxy) {
          ook = 0;
        }
      }
      if (ook) oky = 1;
    }
    l = 1e9, r = yr.size() - 1;
    for (int i = 0; i < yr.size(); i++) {
      if (yr[i] == Maxy) {
        l = min(l, i);
        r = max(r, i);
      }
    }
    if (l != 0 && yr[l - 1] != Miny) {
      int ook = 1;
      for (int i = l; i <= r; i++) {
        if (yr[i] == Miny) {
          ook = 0;
        }
      }
      if (ook) oky = 1;
    }
    printf("%lld\n", min((Maxx - Minx + 1 - okx) * (Maxy - Miny + 1),
                         (Maxx - Minx + 1) * (Maxy - Miny + 1 - oky)));
  }
  return 0;
}
