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;
inline long long MIN(long long x, long long y) { return x < y ? x : y; }
inline long long MAX(long long x, long long y) { return x < y ? y : x; }
struct point {
  long long x, y;
  point &operator=(const point &px) {
    x = px.x;
    y = px.y;
    return *this;
  }
  point operator+(const point &px) const { return (point){x + px.x, y + px.y}; }
  point &operator+=(const point &px) { return *this = *this + px; }
};
struct rct {
  point x, y;
  rct &operator=(const rct &rx) {
    x = rx.x;
    y = rx.y;
    return *this;
  }
  rct operator+(const point &px) const { return (rct){x + px, y + px}; }
  rct operator+(const rct &rx) const {
    return (rct){(point){MIN(x.x, rx.x.x), MIN(x.y, rx.x.y)},
                 (point){MAX(y.x, rx.y.x), MAX(y.y, rx.y.y)}};
  }
  rct &operator=(const point &px) {
    x = y = px;
    return *this;
  }
  rct &operator+=(const rct &rx) { return *this = *this + rx; }
};
const long long mxn = 200010;
const point dir[4] = {(point){0, 1}, (point){-1, 0}, (point){0, -1},
                      (point){1, 0}};
long long inf, minf;
long long I, i, j, k;
long long x, y, z, xx, xy, xz, yx, yy, yz;
point px, py, pz;
rct rx, ry, rz, rxx, rxy, rxz;
long long T, m, n, p;
char a[mxn];
long long mtch[128];
rct b[mxn];
rct f[mxn], g[mxn];
long long ans;
int main() {
  scanf("%lld", &T);
  memset(&inf, 0x3f, sizeof(long long));
  memset(&minf, 0xc0, sizeof(long long));
  mtch['W'] = 0;
  mtch['A'] = 1;
  mtch['S'] = 2;
  mtch['D'] = 3;
  for (I = 1; I <= T; I++) {
    scanf("%s", a + 1);
    n = strlen(a + 1);
    px = (point){0, 0};
    rx = px;
    b[0] = rx;
    f[0] = rx;
    for (i = 1; i <= n; i++) {
      px += dir[mtch[a[i]]];
      ry = px;
      b[i] = ry;
      rx += ry;
      f[i] = rx;
    }
    rx = b[n];
    g[n] = rx;
    for (i = n - 1; i; i--) {
      rx += b[i];
      g[i] = rx;
    }
    ans = inf;
    for (i = 1; i <= n; i++) {
      rx = f[i];
      ry = g[i];
      for (j = 0; j < 4; j++) {
        rz = ry + dir[j];
        rxx = rz + rx;
        ans = MIN(ans, (rxx.y.x - rxx.x.x + 1) * (rxx.y.y - rxx.x.y + 1));
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}
