#include<bits/stdc++.h>
#define endl '\n'
#define int long long

using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;

const int N = 2e5 + 5;
int n, m, k, x, y, T;

void add(int x1, int y1, int x2, int y2, int c, vector<vector<int>> &b){
	b[x1][y1] += c;
	b[x2+1][y1] -= c;
	b[x1][y2+1] -= c;
	b[x2+1][y2+1] += c;
}

void solve()
{
	cin >> n >> m >> k;
	vector<vector<int>> g(n + 10, vector<int>(m + 10));
	vector<vector<int>> b(n + 10, vector<int>(m + 10));
	vector<vector<bool>> vis(n + 10, vector<bool>(m + 10));
	string s; cin >> s;
	
	//所有袋鼠都向上移动一步，相当于上下边界向下移动一步；所有袋鼠都向左移动一步，相当于左右边界向右移动一步。
	int U = 1, D = n, L = 1, R = m; //以左上角为(1,1)作一个n*m的矩阵
	for (int i = 0, u = 1, d = n, l = 1, r = m; i < s.size(); 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);
		
	}
	
	if (U > D || R < L){
		if(k == 0){
			cout << n * m << endl;
		} else{
			cout << 0 << endl;
		}
		return;
	} 
	
	int need_to_die = (D - U + 1) * (R - L + 1) - k;
	if (need_to_die < 0){
		cout << 0 << endl; return;
	} else if(need_to_die == 0){
		cout << n * m << endl; return;
	}
	
	add(U, L, D, R, 1, b);
	vis[U][L] = 1; //表示袋鼠矩形已经走过了以(U,L)为左上角的矩形区域
	for(int i = 0, x1 = U, y1 = L, x2 = D, y2 = R; i < s.size(); i ++){
		if(s[i] == 'U') x1 --, x2 --;
		else if(s[i] == 'D') x1 ++, x2 ++;
		else if(s[i] == 'L') y1 --, y2 --;
		else y1 ++, y2 ++;
		if(vis[x1][y1]) continue; //防止重复计算
		else vis[x1][y1] = 1, add(x1, y1, x2, y2, 1, b);
	}
			
	int ans = 0;
	vector<vector<int>> sum(n + 10, vector<int>(m + 10));
	for(int i = U; i <= D; i ++)
		for(int j = L; j <= R; j ++){
			sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + b[i][j];
			if(sum[i][j] == need_to_die) ans ++;
		}
	cout << ans << endl;
}

signed main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	
	cin >> T;
	while(T--) solve();
	
	return 0;
}