#include <bits/stdc++.h>

using namespace std;

#define F(i, a, b) for (int i = a; i < b; i++)
#define FE(i, a, b) for (int i = a; i <= b; i++)
#define FR(i, a, b) for (int i = a; i > b; i--)
#define FRE(i, a, b) for (int i = a; i >= b; i--)
#define ll                                                                     \
  long long // data types used often, but you don't want to type them time by
            // time
#define ull unsigned long long
#define ui unsigned int
#define us unsigned short
#define FOREACH(i, t)                                                          \
  for (auto i = t.begin(); i != t.end(); i++) // traverse an STL data structure
#define ALL(c) (c).begin(), (c).end() // handy for function like "sort()"
#define IND(e, arr) find(begin(arr), end(arr), e) - begin(arr)
#define IOS                                                                    \
  ios_base::sync_with_stdio(0); // to synchronize the input of cin and scanf
#define INF 1001001001
#define PI 3.1415926535897932384626
#define mp make_pair
#define fi first
#define se second
#define pb push_back

typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<pair<int, int>> vii;
typedef vector<vector<int>> vvi;
typedef vector<bool> vb;

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

#define N 1100

int dp[2 * N][2 * N];

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  string dir = "LDRU", s;
  int t, n, m, k;
  cin >> t;
  while (t--) {
    cin >> n >> m >> k;
    ll x = 0, y = 0, l = 0, d = 0, r = 0, u = 0;
    cin >> s;
    for (int i = -n - 1; i <= n; i++) {
      for (int j = -m - 1; j <= m; j++) {
        dp[N + i][N + j] = 0;
      }
    }
    dp[N][N] = 1;
    for (char c : s) {
      int index = dir.find(c);
      x += dx[index], y += dy[index];
      l = max(l, x);
      r = max(r, -x);
      u = max(u, -y);
      d = max(d, y);
      if (abs(x) <= m && abs(y) <= n) {
        dp[N - y][N + x] = 1;
      }
    }
    for (int i = -n; i <= n; i++) {
      for (int j = -m; j <= m; j++) {
        // cout << dp[N + i][N + j] << ' ';
        dp[N + i][N + j] += dp[N + i - 1][N + j] + dp[N + i][N + j - 1] -
                            dp[N + i - 1][N + j - 1];
      }
      // cout << endl;
    }

    // cout << l << ' ' << r << ' ' << u << ' ' << d << endl;

    ll cnt = n * m - (l + r) * n - (u + d) * max(m - l - r, 0ll);
    cnt = max(cnt, 0ll) - k;
    ii tl = {u, l}, br = {n - 1 - d, m - 1 - r};
    // cout << cnt << endl;
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        int tmp;
        if (l + r >= m || u + d >= n)
          tmp = 0;
        else {
          tmp = dp[N + br.first - i][N + br.second - j] -
                dp[N + br.first - i][N + tl.second - 1 - j] -
                dp[N + tl.first - 1 - i][N + br.second - j] +
                dp[N + tl.first - 1 - i][N + tl.second - 1 - j];
        }
        // cout << tmp << ' ';
        //  cout << (n - 1 - d) - i << ' ' << +(m - 1 - r) - j << endl;
        if (tmp == cnt)
          ans++;
        // printf("(%d, %d), (%d, %d): %d\n", br.first - i, br.second - j,
        //        tl.first - i, tl.second - j, tmp);
      }
    }
    cout << ans << '\n';
  }

  return 0;
}