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 INF = 2 * 1e5 + 1;
int v[INF];
int h[INF];
pair<int, long long int> diminui(int a, int arr[]) {
  int mx = 0;
  int mn = 0;
  int pmx = 0;
  int umn = 0;
  for (int i = 0; i < a; i++) {
    if (arr[i] > mx) {
      mx = arr[i];
      pmx = i;
    } else if (arr[i] <= mn) {
      mn = arr[i];
      umn = i;
    }
  }
  return make_pair(umn + 1 < pmx, mx - mn + 1);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    int vp = 1, hp = 1;
    for (int i = 0; i < str.size(); i++) {
      if (str[i] == 'W') {
        v[vp] = v[vp - 1] + 1;
        vp++;
      } else if (str[i] == 'S') {
        v[vp] = v[vp - 1] - 1;
        vp++;
      } else if (str[i] == 'A') {
        h[hp] = h[hp - 1] + 1;
        hp++;
      } else {
        h[hp] = h[hp - 1] - 1;
        hp++;
      }
    }
    pair<int, int> a = diminui(hp, h);
    pair<int, int> w = diminui(vp, v);
    for (int i = 0; i < hp; i++) {
      h[i] *= -1;
    }
    for (int i = 0; i < vp; i++) {
      v[i] *= -1;
    }
    pair<int, int> d = diminui(hp, h);
    pair<int, int> s = diminui(vp, v);
    if (d.second > s.second) {
      if (s.first || w.first) {
        (s.second)--;
      } else if (a.first || d.first) {
        (a.second)--;
      }
    } else {
      if (a.first || d.first) {
        (a.second)--;
      } else if (s.first || w.first) {
        (s.second)--;
      }
    }
    long long int ans = (long long int)a.second * s.second;
    cout << ans << endl;
  }
}
