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>
#pragma comment(linker, "/STACK: 1024000000,1024000000")
using namespace std;
inline int getint() {
  int _x = 0;
  char _tc = getchar();
  while (_tc < '0' || _tc > '9') _tc = getchar();
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x;
}
const int N = 2e5 + 25;
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int pmx[N], pmn[N], smx[N], smn[N], n, sm[N];
char s[N];
map<char, int> m;
int solve(int* d, long long& nw) {
  long long ans = -((long long)(0x3f3f3f3f3f3f3f3f));
  pmx[0] = -((long long)(0x3f3f3f3f3f3f3f3f)),
  pmn[0] = ((long long)(0x3f3f3f3f3f3f3f3f));
  sm[1] = 0;
  for (int i = 1; i <= n + 1; i++) {
    if (i > 1) sm[i] = sm[i - 1] + d[m[s[i - 1]]];
    pmx[i] = max(pmx[i - 1], sm[i]);
    pmn[i] = min(pmn[i - 1], sm[i]);
  }
  ans = nw = pmx[n + 1] - pmn[n + 1] + 1;
  smx[n + 2] = -((long long)(0x3f3f3f3f3f3f3f3f)),
          smn[n + 2] = ((long long)(0x3f3f3f3f3f3f3f3f));
  for (int i = n + 1; i; i--) {
    smx[i] = max(smx[i + 1], sm[i]);
    smn[i] = min(smn[i + 1], sm[i]);
  }
  long long nmx, nmn;
  for (int i = 1; i <= n; i++) {
    nmx = max(pmx[i], max(smx[i + 1], sm[i]) + 1);
    nmn = min(pmn[i], min(smn[i + 1], sm[i]) + 1);
    ans = min(ans, nmx - nmn + 1);
    nmx = max(pmx[i], max(smx[i + 1], sm[i]) - 1);
    nmn = min(pmn[i], min(smn[i + 1], sm[i]) - 1);
    ans = min(ans, nmx - nmn + 1);
  }
  return ans;
}
int main() {
  m['W'] = 0;
  m['S'] = 1;
  m['A'] = 2;
  m['D'] = 3;
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    long long a, b, c, d;
    a = solve(dx, b);
    c = solve(dy, d);
    cout << min(b * d, min(a * d, b * c)) << '\n';
  }
  return 0;
}
