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;
const int mxN = 2e5 + 45, mod = 1e9 + 7;
const long long inf = 2e18 + 43;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
string s;
void solve() {
  cin >> s;
  int n = s.size();
  int h = 0, m = 0;
  int u = 0, d = 0, l = 0, r = 0;
  for (int i = 0; i < n; ++i) {
    h += s[i] == 'W', h -= s[i] == 'S';
    m += s[i] == 'A', m -= s[i] == 'D';
    u = max(u, h), d = min(d, h);
    r = max(r, m), l = min(l, m);
  }
  long long area = (u - d + 1) * 1ll * (r - l + 1);
  {
    vector<int> a;
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'A' || s[i] == 'D') continue;
      if (s[i] == 'W')
        a.push_back(1);
      else
        a.push_back(-1);
    }
    if (!a.empty()) {
      int cur = 0, mx = 0, mn = 0;
      vector<int> p(a.size());
      p[0] = a[0];
      for (int i = 1; i < a.size(); ++i) p[i] = p[i - 1] + a[i];
      vector<int> suffmx, suffmn;
      int curmx = -mod, curmn = mod;
      for (int i = a.size() - 1; i >= 0; i--) {
        curmx = max(curmx, p[i]);
        curmn = min(curmn, p[i]);
        int addmx, addmn;
        if (i) {
          addmx = max(curmx - p[i - 1], curmn - p[i - 1]);
          addmn = min(curmx - p[i - 1], curmn - p[i - 1]);
        } else {
          addmx = curmx;
          addmn = curmn;
        }
        addmx = max(addmx, 0);
        addmn = min(addmn, 0);
        suffmx.push_back(addmx);
        suffmn.push_back(addmn);
      }
      reverse(suffmx.begin(), suffmx.end());
      reverse(suffmn.begin(), suffmn.end());
      int now = 0;
      mx = 0, mn = 0;
      {
        ++now;
        int nmx = max(mx, now + suffmx[0]);
        int nmn = min(mn, now + suffmn[0]);
        --now;
        --now;
        area = min(area, (nmx - nmn + 1) * 1ll * (r - l + 1));
        nmx = max(mx, now + suffmx[0]);
        nmn = min(mn, now + suffmn[0]);
        ++now;
        area = min(area, (nmx - nmn + 1) * 1ll * (r - l + 1));
      }
      for (int i = 0; i < a.size() - 1; ++i) {
        now += a[i];
        mx = max(mx, now);
        mn = min(mn, now);
        {
          ++now;
          int nmx = max(mx, now + suffmx[i + 1]);
          int nmn = min(mn, now + suffmn[i + 1]);
          --now;
          --now;
          area = min(area, (nmx - nmn + 1) * 1ll * (r - l + 1));
          nmx = max(mx, now + suffmx[i + 1]);
          nmn = min(mn, now + suffmn[i + 1]);
          ++now;
          area = min(area, (nmx - nmn + 1) * 1ll * (r - l + 1));
        }
      }
    }
  }
  {
    vector<int> a;
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'W' || s[i] == 'S') continue;
      if (s[i] == 'D')
        a.push_back(1);
      else
        a.push_back(-1);
    }
    if (!a.empty()) {
      vector<int> p(a.size());
      p[0] = a[0];
      for (int i = 1; i < a.size(); ++i) p[i] = p[i - 1] + a[i];
      int cur = 0, mx = 0, mn = 0;
      vector<int> suffmx, suffmn;
      int curmx = -mod, curmn = mod;
      for (int i = a.size() - 1; i >= 0; i--) {
        curmx = max(curmx, p[i]);
        curmn = min(curmn, p[i]);
        int addmx, addmn;
        if (i) {
          addmx = max(curmx - p[i - 1], curmn - p[i - 1]);
          addmn = min(curmx - p[i - 1], curmn - p[i - 1]);
        } else {
          addmx = curmx;
          addmn = curmn;
        }
        addmx = max(addmx, 0);
        addmn = min(addmn, 0);
        suffmx.push_back(addmx);
        suffmn.push_back(addmn);
      }
      reverse(suffmx.begin(), suffmx.end());
      reverse(suffmn.begin(), suffmn.end());
      int now = 0;
      mx = 0, mn = 0;
      {
        ++now;
        int nmx = max(mx, now + suffmx[0]);
        int nmn = min(mn, now + suffmn[0]);
        --now;
        --now;
        area = min(area, (nmx - nmn + 1) * 1ll * (u - d + 1));
        nmx = max(mx, now + suffmx[0]);
        nmn = min(mn, now + suffmn[0]);
        ++now;
        area = min(area, (nmx - nmn + 1) * 1ll * (u - d + 1));
      }
      for (int i = 0; i < a.size() - 1; ++i) {
        now += a[i];
        mx = max(mx, now);
        mn = min(mn, now);
        {
          ++now;
          int nmx = max(mx, now + suffmx[i + 1]);
          int nmn = min(mn, now + suffmn[i + 1]);
          --now;
          --now;
          area = min(area, (nmx - nmn + 1) * 1ll * (u - d + 1));
          nmx = max(mx, now + suffmx[i + 1]);
          nmn = min(mn, now + suffmn[i + 1]);
          ++now;
          area = min(area, (nmx - nmn + 1) * 1ll * (u - d + 1));
        }
      }
    }
  }
  cout << area;
}
int main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) solve(), cout << '\n';
  return 0;
}
