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 maxn = 3e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b > 0 ? gcd(b, a % b) : a; }
int n, m, T;
char s[maxn];
struct node {
  long long l, r, d, w;
} u[maxn], v[maxn];
int dir[4][4] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int main() {
  cin >> T;
  while (T--) {
    cin >> s + 1;
    n = strlen(s + 1);
    int lx = 0, rx = 0, ly = 0, ry = 0;
    int x = 0, y = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'D')
        x++;
      else if (s[i] == 'A')
        x--;
      else if (s[i] == 'W')
        y++;
      else
        y--;
      lx = min(lx, x);
      rx = max(rx, x);
      ly = min(ly, y);
      ry = max(ry, y);
      u[i] = {lx, rx, ly, ry};
    }
    lx = rx = x, ly = ry = y;
    for (int i = n; i >= 1; i--) {
      if (s[i] == 'D')
        x--;
      else if (s[i] == 'A')
        x++;
      else if (s[i] == 'W')
        y--;
      else
        y++;
      lx = min(lx, x);
      rx = max(rx, x);
      ly = min(ly, y);
      ry = max(ry, y);
      v[i] = {lx, rx, ly, ry};
    }
    u[0] = {0, 0, 0, 0};
    long long ans = 1e18;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 4; j++) {
        lx = min(u[i - 1].l, v[i].l + dir[j][0]);
        rx = max(u[i - 1].r, v[i].r + dir[j][0]);
        ly = min(u[i - 1].d, v[i].d + dir[j][1]);
        ry = max(u[i - 1].w, v[i].w + dir[j][1]);
        ans = min(ans, 1ll * abs(rx - lx + 1) * abs(ry - ly + 1));
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
