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;
using ll = long long;
template <class T>
void chkmin(T &x, const T &y) {
  if (y < x) {
    x = y;
  }
}
template <class T>
void chkmax(T &x, const T &y) {
  if (y > x) {
    x = y;
  }
}
template <class T>
void sort(vector<T> &a) {
  sort(a.begin(), a.end());
}
template <class T, class Cmp>
void sort(vector<T> &a, Cmp cmp) {
  sort(a.begin(), a.end(), cmp);
}
template <class T>
void unique(vector<T> &a) {
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
int readInt() {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x;
}
string readString() {
  string s;
  char c = getchar();
  while (isspace(c)) {
    c = getchar();
  }
  while (!isspace(c)) {
    s += c;
    c = getchar();
  }
  return s;
}
void solve_single_test(int test_id = 1) {
  string s = readString();
  int n = s.length();
  vector<int> preL(n), preR(n), preU(n), preD(n);
  vector<int> sufL(n), sufR(n), sufU(n), sufD(n);
  int L = 0, R = 0, U = 0, D = 0, x = 0, y = 0;
  for (int i = 0; i < n; ++i) {
    switch (s[i]) {
      case 'W':
        ++y;
        break;
      case 'A':
        --x;
        break;
      case 'S':
        --y;
        break;
      case 'D':
        ++x;
        break;
    }
    chkmax(R, x);
    chkmin(L, x);
    chkmax(U, y);
    chkmin(D, y);
    preL[i] = L - x;
    preR[i] = R - x;
    preU[i] = U - y;
    preD[i] = D - y;
  }
  L = 0;
  R = 0;
  U = 0;
  D = 0;
  x = 0;
  y = 0;
  for (int i = n - 1; i >= 0; --i) {
    switch (s[i]) {
      case 'W':
        --y;
        break;
      case 'A':
        ++x;
        break;
      case 'S':
        ++y;
        break;
      case 'D':
        --x;
        break;
    }
    chkmax(R, x);
    chkmin(L, x);
    chkmax(U, y);
    chkmin(D, y);
    sufL[i] = L - x;
    sufR[i] = R - x;
    sufU[i] = U - y;
    sufD[i] = D - y;
  }
  ll ans = 1ll * (sufR[0] - sufL[0] + 1) * (sufU[0] - sufD[0] + 1);
  for (int i = 0; i < n - 1; ++i) {
    chkmin(ans,
           1ll *
               (max(preR[i], sufR[i + 1] + 1) - min(preL[i], sufL[i + 1] + 1) +
                1) *
               (max(preU[i], sufU[i + 1]) - min(preD[i], sufD[i + 1]) + 1));
    chkmin(ans,
           1ll *
               (max(preR[i], sufR[i + 1] - 1) - min(preL[i], sufL[i + 1] - 1) +
                1) *
               (max(preU[i], sufU[i + 1]) - min(preD[i], sufD[i + 1]) + 1));
    chkmin(ans,
           1ll * (max(preR[i], sufR[i + 1]) - min(preL[i], sufL[i + 1]) + 1) *
               (max(preU[i], sufU[i + 1] + 1) - min(preD[i], sufD[i + 1] + 1) +
                1));
    chkmin(ans,
           1ll * (max(preR[i], sufR[i + 1]) - min(preL[i], sufL[i + 1]) + 1) *
               (max(preU[i], sufU[i + 1] - 1) - min(preD[i], sufD[i + 1] - 1) +
                1));
  }
  printf("%I64d\n", ans);
}
void solve_multiple_tests() {
  int n_tests;
  scanf("%d", &n_tests);
  for (int i = 1; i <= n_tests; ++i) {
    solve_single_test(i);
  }
}
int main() {
  solve_multiple_tests();
  return 0;
}
