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;
char s[200005];
int sum[2][200005];
int mm[4][200005];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    for (int i = 1; i <= n; i++) {
      sum[0][i] = sum[0][i - 1];
      sum[1][i] = sum[1][i - 1];
      if (s[i] == 'W') {
        sum[0][i]++;
      } else if (s[i] == 'S') {
        sum[0][i]--;
      } else if (s[i] == 'A') {
        sum[1][i]--;
      } else {
        sum[1][i]++;
      }
    }
    mm[0][n] = mm[1][n] = sum[0][n];
    mm[2][n] = mm[3][n] = sum[1][n];
    int xmi = min(0, sum[1][n]), ymi = min(0, sum[0][n]),
        xma = max(0, sum[1][n]), yma = max(0, sum[0][n]);
    for (int i = n - 1; i >= 0; i--) {
      xmi = min(xmi, sum[1][i]);
      ymi = min(ymi, sum[0][i]);
      xma = max(xma, sum[1][i]);
      yma = max(yma, sum[0][i]);
      mm[0][i] = min(sum[0][i], mm[0][i + 1]);
      mm[1][i] = max(sum[0][i], mm[1][i + 1]);
      mm[2][i] = min(sum[1][i], mm[2][i + 1]);
      mm[3][i] = max(sum[1][i], mm[3][i + 1]);
    }
    int w = xma - xmi + 1, h = yma - ymi + 1;
    int ans1 = 2000000, ans2 = 2000000;
    long long ans = 1e18;
    xmi = ymi = xma = yma = 0;
    for (int i = 1; i <= n; i++) {
      xmi = min(xmi, sum[1][i]);
      ymi = min(ymi, sum[0][i]);
      xma = max(xma, sum[1][i]);
      yma = max(yma, sum[0][i]);
      ans1 = 2000000, ans2 = 2000000;
      ans2 = min(ans2, max(xma, mm[3][i] - 1) - min(xmi, mm[2][i] - 1) + 1);
      ans1 = min(ans1, max(yma, mm[1][i] - 1) - min(ymi, mm[0][i] - 1) + 1);
      ans2 = min(ans2, max(xma, mm[3][i] + 1) - min(xmi, mm[2][i] + 1) + 1);
      ans1 = min(ans1, max(yma, mm[1][i] + 1) - min(ymi, mm[0][i] + 1) + 1);
      ans = min(ans, min((long long)ans1 * w, (long long)ans2 * h));
    }
    printf("%lld\n", ans);
  }
  return 0;
}
