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 ri() {
  int x;
  if (scanf(" %d", &x) == EOF) return -1;
  return x;
}
long long int rl() {
  long long int x;
  if (cin >> x) return x;
  return -1;
}
string rs() {
  string s;
  if (cin >> s) return s;
  return "";
}
struct SingleCase {
  string s;
  bool solveCase() {
    s = rs();
    if (s == "") return false;
    for (char& c : s) {
      if (c == 'W') c = '0';
      if (c == 'S') c = '1';
      if (c == 'A') c = '2';
      if (c == 'D') c = '3';
    }
    long long int a1 = getSpan(s, {-1, 1, 0, 0});
    long long int a2 = getSpan(s, {1, -1, 0, 0});
    long long int a3 = getSpan(s, {0, 0, -1, 1});
    long long int a4 = getSpan(s, {0, 0, 1, -1});
    long long int b1 = getMinSpan(a1, a2);
    long long int b3 = getMinSpan(a3, a4);
    long long int ret = min(b1 * max(a3, a4), b3 * max(a1, a2));
    ret = min(ret, max(a1, a2) * max(a3, a4));
    cout << ret << endl;
    return true;
  }
  long long int getMinSpan(int a1, int a2) {
    int ret = max(a1, a2);
    if (a1 > 2) {
      ret = min(ret, max(a1 - 1, a2));
    }
    if (a2 > 2) {
      ret = min(ret, max(a1, a2 - 1));
    }
    return ret;
  }
  int getSpan(string& s, array<int, 4> d) {
    int t = 0;
    int ret = 0;
    for (char& c : s) {
      int x = d[c - '0'];
      t += x;
      ret = max(ret, t);
      if (t < 0) t = 0;
    }
    return ret + 1;
  }
};
int main() {
  ri();
  while (SingleCase().solveCase()) {
  };
  return 0;
}
