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 = 2e6 + 5;
char s[N];
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> s;
    int n = strlen(s);
    int l = 0;
    int r = 0;
    int f = 0;
    int b = 0;
    int suml1 = 0;
    int sumr1 = 0;
    int sumf1 = 0;
    int sumb1 = 0;
    int now1 = 0;
    int now2 = 0;
    int posl = 0;
    int posr = 0;
    int posf = 0;
    int posb = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'W') {
        now2++;
        if (f < now2) {
          f = now2;
          posf = i;
        }
        sumf1++;
      }
      if (s[i] == 'S') {
        now2--;
        if (b > now2) {
          b = now2;
          posb = i;
        }
        sumb1++;
      }
      if (s[i] == 'A') {
        now1--;
        if (l > now1) {
          l = now1;
          posl = i;
        }
        suml1++;
      }
      if (s[i] == 'D') {
        now1++;
        if (r < now1) {
          r = now1;
          posr = i;
        }
        sumr1++;
      }
    }
    int suml2 = 1;
    int sumr2 = -1;
    int sumf2 = -1;
    int sumb2 = 1;
    now1 = 0;
    now2 = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'W') {
        now2++;
      }
      if (s[i] == 'S') {
        now2--;
      }
      if (s[i] == 'A') {
        now1--;
      }
      if (s[i] == 'D') {
        now1++;
      }
      if (i > posl) {
        sumr2 = max(sumr2, now1);
      }
      if (i > posr) {
        suml2 = min(suml2, now1);
      }
      if (i > posf) {
        sumb2 = min(sumb2, now2);
      }
      if (i > posb) {
        sumf2 = max(sumf2, now2);
      }
    }
    int ans1 = r - l + 1;
    int ans2 = f - b + 1;
    long long ans = (long long)ans1 * (long long)ans2;
    if (l < 0 && r > 0) {
      if (l < 0 && sumr2 < r) {
        ans = min(ans, (long long)max(ans1 - 1, 2) * ans2);
      }
      if (r > 0 && suml2 > l) {
        ans = min(ans, (long long)max(ans1 - 1, 2) * ans2);
      }
    } else {
      if (suml1 != sumr1) {
        if (l < 0 && sumr2 < r) {
          ans = min(ans, (long long)max(ans1 - 1, 2) * ans2);
        }
        if (r > 0 && suml2 > l) {
          ans = min(ans, (long long)max(ans1 - 1, 2) * ans2);
        }
      }
    }
    if (f > 0 && b < 0) {
      if (b < 0 && sumf2 < f) {
        ans = min(ans, (long long)max(ans2 - 1, 2) * ans1);
      }
      if (f > 0 && sumb2 > b) {
        ans = min(ans, (long long)max(ans2 - 1, 2) * ans1);
      }
    } else {
      if (sumf1 != sumb1) {
        if (b < 0 && sumf2 < f) {
          ans = min(ans, (long long)max(ans2 - 1, 2) * ans1);
        }
        if (f > 0 && sumb2 > b) {
          ans = min(ans, (long long)max(ans2 - 1, 2) * ans1);
        }
      }
    }
    cout << ans << "\n";
  }
}
