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;
  cin >> T;
  while (T--) {
    char str[200005];
    scanf("%s", str);
    int len = strlen(str);
    long long leftN = 0, rightN = 0, ln = 0, rn = 0;
    long long ln1 = 0, ln2 = 0, rn1 = 0, rn2 = 0;
    int lflag = 0, rflag = 0, uflag = 0, dflag = 0;
    for (int i = 0; i < len; i++) {
      switch (str[i]) {
        case 'W':
          ln++;
          break;
        case 'S':
          ln--;
          break;
        case 'A':
          rn++;
          break;
        case 'D':
          rn--;
          break;
      }
      if (ln == 0) {
        if (ln1 == ln) {
          lflag = 1;
        }
        if (ln2 == ln) {
          rflag = 1;
        }
      } else if (ln > 0) {
        if (ln > ln1) {
          ln1 = ln;
          lflag = 1;
          rflag = 0;
        } else if (ln1 == ln) {
          lflag = 1;
        }
      } else {
        if (abs(ln) > ln2) {
          ln2 = abs(ln);
          rflag = 1;
          lflag = 0;
        } else if (ln2 == abs(ln)) {
          rflag = 1;
        }
      }
      if (rn == 0) {
        if (rn1 == rn) {
          uflag = 1;
        }
        if (rn2 == rn) {
          dflag = 1;
        }
      } else if (rn > 0) {
        if (rn > rn1) {
          rn1 = rn;
          uflag = 1;
          dflag = 0;
        } else if (rn1 == rn) {
          uflag = 1;
        }
      } else {
        if (abs(rn) > rn2) {
          rn2 = abs(rn);
          dflag = 1;
          uflag = 0;
        } else if (rn2 == abs(rn)) {
          dflag = 1;
        }
      }
    }
    leftN = ln1 + ln2 + 1;
    rightN = rn1 + rn2 + 1;
    long long ans = leftN * rightN;
    if ((leftN > 2) && (lflag * rflag == 0)) {
      ans = min(ans, (leftN - 1) * rightN);
    }
    if ((rightN > 2) && (uflag * dflag == 0)) {
      ans = min(ans, leftN * (rightN - 1));
    }
    printf("%lld\n", ans);
  }
  return 0;
}
