#include <bits/stdc++.h>
#define MAXN 1000
#define MAXM 1000
#define MAXS ((int) 1e6)
using namespace std;

int n, m, K, ans;
char s[MAXS + 10];

int f[MAXN * 2 + 10][MAXM * 2 + 10];

void solve() {
    scanf("%d%d%d%s", &n, &m, &K, s + 1);
    // 维护边界移动的最值
    int U = 1, D = n, L = 1, R = m;
    for (int i = 1, u = 1, d = n, l = 1, r = m; s[i]; i++) {
        if (s[i] == 'U') u++, d++;
        else if (s[i] == 'D') u--, d--;
        else if (s[i] == 'L') l++, r++;
        else l--, r--;
        U = max(U, u);
        D = min(D, d);
        L = max(L, l);
        R = min(R, r);
    }

    // 特判没有任何袋鼠存留的情况
    if (U > D || L > R) {
        if (K == 0) printf("%d\n", n * m);
        else printf("0\n");
        return;
    }

    // 计算还差几只袋鼠需要用洞解决
    int delta = (D - U + 1) * (R - L + 1) - K;
    if (delta < 0) {
        printf("0\n");
        return;
    }

    // 记录洞的偏移量
    // 偏移量范围是 [-n, n] 以及 [-m, m]，因此都加上 n + 1 和 m + 1 变成非负数
    for (int i = 0; i <= n * 2 + 5; i++) for (int j = 0; j <= m * 2 + 5; j++) f[i][j] = 0;
    int BIAS_R = n + 1, BIAS_C = m + 1;
    f[BIAS_R][BIAS_C] = 1;
    for (int i = 1, r = BIAS_R, c = BIAS_C; s[i]; i++) {
        if (s[i] == 'U') r++;
        else if (s[i] == 'D') r--;
        else if (s[i] == 'L') c++;
        else if (s[i] == 'R') c--;
        f[r][c] = 1;
    }
    // 二维前缀和
    for (int i = 1; i <= n * 2 + 5; i++) for (int j = 1; j <= m * 2 + 5; j++) f[i][j] += f[i][j - 1];
    for (int i = 1; i <= n * 2 + 5; i++) for (int j = 1; j <= m * 2 + 5; j++) f[i][j] += f[i - 1][j];

    ans = 0;
    // 枚举洞的初始坐标
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
        int biasR = BIAS_R - i, biasC = BIAS_C - j;
        int t = f[D + biasR][R + biasC] - f[U - 1 + biasR][R + biasC] - f[D + biasR][L - 1 + biasC] + f[U - 1 + biasR][L - 1 + biasC];
        if (t == delta) ans++;
    }
    printf("%d\n", ans);
}

int main() {
    int tcase; scanf("%d", &tcase);
    while (tcase--) solve();
    return 0;
}