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;
template <class T>
inline void read(T &x) {
  x = 0;
  int c = getchar(), f = 1;
  for (; !isdigit(c); c = getchar())
    if (c == 45) f = -1;
  for (; isdigit(c); c = getchar()) (x *= 10) += f * (c - '0');
}
int n;
char s[200000 + 5];
vector<int> seq[2];
int pos[200000 + 5], pre[2][200000 + 5], suf[2][200000 + 5];
int calc(vector<int> vec) {
  int l = 0, r = 0, p = 0;
  for (int x : vec) p += x, l = min(l, p), r = max(r, p);
  return r - l + 1;
}
int insert(vector<int> vec) {
  pos[0] = pre[0][0] = pre[1][0] = 0;
  suf[0][vec.size()] = suf[1][vec.size()] = 0;
  for (int i = 0; i < (int)vec.size(); i++)
    pos[i + 1] = pos[i] + vec[i], pre[0][i + 1] = min(pre[0][i], pos[i + 1]),
            pre[1][i + 1] = max(pre[1][i], pos[i + 1]);
  for (int i = (int)vec.size() - 1; ~i; i--)
    suf[0][i] = min(suf[0][i + 1] + vec[i], 0),
    suf[1][i] = max(suf[1][i + 1] + vec[i], 0);
  int mi = 0x3f3f3f3f;
  for (int d = -1; d <= 1; d += 2)
    for (int i = 0, l, r; i <= (int)vec.size(); i++)
      l = min(pre[0][i], pos[i] + suf[0][i] + d),
      r = max(pre[1][i], pos[i] + suf[1][i] + d), mi = min(mi, r - l + 1);
  return mi;
}
int main() {
  int T;
  read(T);
  while (T--) {
    scanf("%s", s), n = (int)strlen(s);
    for (int i = 0; i < n; i++) {
      if (s[i] == 'W') seq[0].push_back(1);
      if (s[i] == 'S') seq[0].push_back(-1);
      if (s[i] == 'D') seq[1].push_back(1);
      if (s[i] == 'A') seq[1].push_back(-1);
    }
    long long a = 1LL * insert(seq[0]) * calc(seq[1]);
    long long b = 1LL * calc(seq[0]) * insert(seq[1]);
    long long c = 1LL * calc(seq[0]) * calc(seq[1]);
    printf("%lld\n", min(min(a, b), c));
    seq[0].clear(), seq[1].clear();
  }
  return 0;
}
