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 GCC optimize("Ofast,no-stack-protector,unroll-loops", \
                     "omit-frame-pointer", "inline")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,fma,tune=native")
#pragma GCC option("arch=native", "no-zero-upper")
using namespace std;
const long long mod = (long long)(1e9 + 7);
const long double eps = (1e-9);
const long long oo = (long long)(1e18);
long long solve(vector<int>& ac, long long w) {
  vector<int> mi_ini;
  vector<int> ma_ini;
  vector<int> mi_fin;
  vector<int> ma_fin;
  int n = ac.size();
  mi_ini.push_back(0);
  ma_ini.push_back(0);
  mi_fin.push_back(ac.back());
  ma_fin.push_back(ac.back());
  for (int i = 1; i < n; i++) {
    mi_ini.push_back(min(mi_ini.back(), ac[i]));
    ma_ini.push_back(max(ma_ini.back(), ac[i]));
  }
  for (int i = n - 2; i >= 0; i--) {
    mi_fin.push_back(min(mi_fin.back(), ac[i]));
    ma_fin.push_back(max(ma_fin.back(), ac[i]));
  }
  reverse((mi_fin).begin(), (mi_fin).end());
  reverse((ma_fin).begin(), (ma_fin).end());
  int res = ma_ini.back() - mi_ini.back() + 1;
  for (int i = 1; i < n - 1; i++) {
    int ma1 = max(ma_ini[i], ma_fin[i + 1] + 1);
    int mi1 = min(mi_ini[i], mi_fin[i + 1] + 1);
    ma1 = max(ma1, ac[i] + 1);
    res = min(res, ma1 - mi1 + 1);
    int ma2 = max(ma_ini[i], ma_fin[i + 1] - 1);
    int mi2 = min(mi_ini[i], mi_fin[i + 1] - 1);
    mi2 = min(mi2, ac[i] - 1);
    res = min(res, ma2 - mi2 + 1);
  }
  return w * res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string second;
    cin >> second;
    vector<int> ach;
    vector<int> acv;
    ach.push_back(0);
    acv.push_back(0);
    int n = second.size();
    for (int i = 0; i < n; i++) {
      int c = second[i];
      if (c == 'W') acv.push_back(acv.back() + 1);
      if (c == 'S') acv.push_back(acv.back() - 1);
      if (c == 'D') ach.push_back(ach.back() + 1);
      if (c == 'A') ach.push_back(ach.back() - 1);
    }
    int mih = 0;
    int mah = 0;
    int miv = 0;
    int mav = 0;
    for (auto v : ach) {
      mih = min(mih, v);
      mah = max(mah, v);
    }
    for (auto v : acv) {
      miv = min(miv, v);
      mav = max(mav, v);
    }
    long long dh = mah - mih + 1;
    long long dv = mav - miv + 1;
    cout << (min(solve(ach, dv), solve(acv, dh))) << '\n';
  }
  return 0;
}
