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;
char s[2000005];
vector<long long> vmix, vmiy, vmx, vmy;
int main() {
  long long valx, valy, mix, miy, mx, my, nrmix, nrmiy, nrmx, nrmy, x, y, t, ok,
      i, v1, v2, v3, v4;
  ios_base::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> s;
    x = y = 0;
    mix = miy = mx = my = 0;
    nrmix = nrmiy = nrmx = nrmy = 1;
    vmix.clear();
    vmiy.clear();
    vmx.clear();
    vmy.clear();
    vmix.push_back(0);
    vmiy.push_back(0);
    vmx.push_back(0);
    vmy.push_back(0);
    for (i = 0; s[i] != NULL; i++) {
      switch (s[i]) {
        case ('W'):
          x++;
          break;
        case ('A'):
          y--;
          break;
        case ('S'):
          x--;
          break;
        case ('D'):
          y++;
          break;
      }
      if (mix > x) {
        mix = x;
        vmix.clear();
      };
      if (miy > y) {
        miy = y;
        vmiy.clear();
      };
      if (mx < x) {
        mx = x;
        vmx.clear();
      };
      if (my < y) {
        my = y;
        vmy.clear();
      };
      if (mix == x) vmix.push_back(i);
      if (miy == y) vmiy.push_back(i);
      if (mx == x) vmx.push_back(i);
      if (my == y) vmy.push_back(i);
    }
    valx = (mx - mix + 1);
    valy = (my - miy + 1);
    ok = 0;
    if (valx <= valy && valx > 2) {
      v1 = vmix[0];
      v2 = vmix[vmix.size() - 1];
      v3 = vmx[0];
      v4 = vmx[vmx.size() - 1];
      if (v4 < v1 || v2 < v3) {
        ok = 1;
        valx--;
      }
    }
    v1 = vmiy[0];
    v2 = vmiy[vmiy.size() - 1];
    v3 = vmy[0];
    v4 = vmy[vmy.size() - 1];
    if (!ok && valy > 2)
      if (v4 < v1 || v2 < v3) valy--, ok = 1;
    v1 = vmix[0];
    v2 = vmix[vmix.size() - 1];
    v3 = vmx[0];
    v4 = vmx[vmx.size() - 1];
    if (!ok && valx > 2)
      if (v4 < v1 || v2 < v3) {
        ok = 1;
        valx--;
      }
    cout << valx * valy << '\n';
  }
  return 0;
}
