#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#define int long long

using namespace std;

constexpr int N = 1005;

string op;

int n, m, k;
int U, D, L, R, U_, D_, L_, R_;
bool st[N][N];
int g[N][N];

void add(int x1, int y1, int x2, int y2){
    if (st[x1][y1])
        return;
    st[x1][y1] = true;
    g[x1][y1]++, g[x2 + 1][y1]--, g[x1][y2 + 1]--, g[x2 + 1][y2 + 1]++;
}

void solve() {
    cin >> n >> m >> k >> op;
    U_ = L_ = U = L = 1, R_ = R = m, D_ = D = n;
    memset(st, false, sizeof st), memset(g, 0, sizeof g);
    //确定边界
    for (auto i: op){
        if (i == 'L')
            L_++, R_++;
        else if (i == 'R')
            L_--, R_--;
        else if (i == 'U')
            U_++, D_++;
        else
            U_--, D_--;
        L = max(L, L_), R = min(R, R_), U = max(U, U_), D = min(D, D_);
    }
    //无袋鼠剩余情况
    if (U > D || L > R){
        if (k)
            puts("0");
        else
            cout << n * m << endl;
        return;
    }
    //统计袋鼠经过格子的情况
    int x = (D - U + 1) * (R - L + 1), cnt = 0;
    add(U, L, D, R);
    for (auto i: op){
        if (i == 'L')
            L--, R--;
        else if (i == 'R')
            L++, R++;
        else if (i == 'U')
            U--, D--;
        else
            U++, D++;
        add(U, L, D, R);
    }
    //二位前缀和+统计答案
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            g[i][j] += g[i - 1][j] + g[i][j - 1] - g[i - 1][j - 1];

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            if (x - g[i][j] == k)
                 cnt++;
    cout << cnt << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int _;
    cin >> _;

    while (_ --) {
        solve();
    }

    return 0;
}