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:10000000000")
using namespace std;
const int MOD = 1000000007;
const int INF = 1e5;
const long long INF2 = 1000000007LL * 1000000007LL;
const long double EPS = 1e-9;
const int SIZE = 300100;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1000000000, 1000000000);
int tt;
string a;
long long xl = 0, xr = 0, yl = 0, yr = 0, x = 0, y = 0;
int ixl, ixr, iyl, iyr;
long long fn(string& a) {
  xl = 0, xr = 0, yl = 0, yr = 0, x = 0, y = 0;
  ixl = ixr = iyl = iyr = -1;
  for (int i = 0; i < a.size(); ++i) {
    switch (a[i]) {
      case 'W':
        ++y;
        break;
      case 'S':
        --y;
        break;
      case 'A':
        --x;
        break;
      case 'D':
        ++x;
        break;
    }
    if (xl > x) {
      xl = x;
      ixl = i;
    }
    if (xr < x) {
      xr = x;
      ixr = i;
    }
    if (yl > y) {
      yl = y;
      iyl = i;
    }
    if (yr < y) {
      yr = y;
      iyr = i;
    }
  }
  return (llabs(xr - xl) + 1) * (llabs(yr - yl) + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> tt;
  for (int rep_iter = 0; rep_iter < tt; ++rep_iter) {
    cin >> a;
    long long s = fn(a);
    vector<string> v;
    if (xr > 0) {
      v.push_back(a.substr(0, ixr) + string("A") + a.substr(ixr));
    }
    if (xl < 0) {
      v.push_back(a.substr(0, ixl) + string("D") + a.substr(ixl));
    }
    if (yr > 0) {
      v.push_back(a.substr(0, iyr) + string("S") + a.substr(iyr));
    }
    if (yl < 0) {
      v.push_back(a.substr(0, iyl) + string("W") + a.substr(iyl));
    }
    for (auto& i : v) {
      s = min(s, fn(i));
    }
    cout << s << '\n';
  }
  return 0;
}
