#include<bits/stdc++.h>
#define int long long
#define mod 1000000007
using namespace std;
const int N = 2005;
int n, m, k;
string s;
int xl, xr, yl, yr;

vector<pair<int, int> > nums;

int mp[N][N];

void solve() {
    cin >> n >> m >> k;
    for (int i = 0; i <= 2 * n; i++) {
        for (int j = 0; j <= 2 * m; j++) {
            mp[i][j] = 0;
        }
    }
    cin >> s;
    int px = 0, py = 0;
    int disux = 0, disdx = 0, disly = 0,disry = 0;
    for (int i = 0; i < s.length(); i++) {
        if (s[i] == 'U') {
            px--;
        }
        else if (s[i] == 'D') {
            px++;
        }
        else if (s[i] == 'L') {
            py--;
        }
        else if (s[i] == 'R') {
            py++;
        }
        //判定位置，然后更新dis，然后判定k=n*m,k=0的情况
        disux = min(disux, px); disdx = max(disdx, px);
        disly = min(disly, py); disry = max(disry, py);
    }
    xl = 1 - disux; yl = 1 - disly;
    xr = n - disdx; yr = m - disry;
    int sum = (yr - yl + 1) * (xr - xl + 1);
    int x = 0, y = 0;
    nums.push_back({ x,y });
    for (int i = 0; i < s.length(); i++) {
        if (s[i] == 'U') {
            x--;
            nums.push_back({ x,y });
        }
        else if (s[i] == 'D') {
            x++;
            nums.push_back({ x,y });
        }
        else if (s[i] == 'L') {
            y--;
            nums.push_back({ x,y });
        }
        else if (s[i] == 'R') {
            y++;
            nums.push_back({ x,y });
        }
    }

      
    int mix = 1e18, miy = 1e18;
    for (auto p = nums.begin(); p != nums.end(); p++) {
        mix = min(mix, p->first);
        miy = min(miy, p->second);
    }
    mix--; miy--;
    mix = -mix, miy = -miy;
    for (auto p = nums.begin(); p != nums.end(); p++) {
        mp[p->first + mix][p->second + miy] = 1;
    }
        
    for (int i = 1; i <= 2 * n; i++) {
        for (int j = 1; j <= 2 * m; j++) {
            mp[i][j] = mp[i][j] + mp[i - 1][j] + mp[i][j - 1] - mp[i - 1][j - 1];
            //cout<<mp[i][j]<<' ';
        }//cout<<endl;
    }
    //cout<<sum<<endl;
    int ans = 0;
    //枚举起点，1,1一直到n,m;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            //枚举在原矩阵中起点的位置，获取位置关系
            ///右下，约束为最大n+mix,m+miy
            //左上，如果越界，约束为1,1
            int dx = (xl - i), dy = (yl -j);
            int x = mix + dx, y = miy + dy;
            int xx = mix + dx + (xr - xl), yy = miy + dy + (yr - yl);
            if (xx <= 0 || yy <= 0) {
                if (0 == sum - k)ans++;
                continue;
            }
            x = max(1ll, x), y = max(1ll, y);
            if ((mp[xx][yy] + mp[x - 1][y - 1] - mp[x - 1][yy] - mp[xx][y - 1]) == sum - k)ans++;

          //  int xx = i, yy = j;
           // int x = i - (xr - xl), y = j - (yr - yl);
          //  x = max(1ll, x); y = max(1ll, y);
            //if ((mp[xx][yy] + mp[x - 1][y - 1] - mp[x - 1][yy] - mp[xx][y - 1]) == sum - k)ans++;


            //放左上角
        }
    }

    cout << ans << endl;
    return;
}
signed main() {
    //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T_case = 1;
    cin >> T_case;
    while (T_case--) {
        solve();
    }
    return 0;
}