#include<iostream>
using namespace std;
#include<vector>
#include<string>
#include<map>

map<char, int> mp =
{
	{'U', 0},
	{'D', 1},
	{'L', 2},
	{'R', 3}
};

vector<int> dx = {1, -1, 0, 0}, dy = {0, 0, 1, -1};

void solve(){
	int n, m, k;
	cin >> n >> m >> k;

	string s;
	cin >> s;

	vector<vector<int>> pre(2 * n, vector<int>(2 * m));
	int t = 0, d = 0, l = 0, r = 0;

	int bx = n, by = m;
	int tx = 0, ty = 0;
	pre[bx][by] = 1;

	for(int i = 0; i < s.size(); ++i){
		bx += dx[mp[s[i]]], by += dy[mp[s[i]]];
		if(bx >= 1 && bx <= 2 * n - 1 && by >= 1 && by <= 2 * m - 1) pre[bx][by] = 1;

		tx += dx[mp[s[i]]], ty += dy[mp[s[i]]];
		t = max(t, -tx), d = max(d, tx);
		l = max(l, -ty), r = max(r, ty);
	}

	if(t + d >= n || l + r >= m){
		if(k == 0){
			cout << n * m << "\n";
		}
		else cout << 0 << "\n";
		return;
	}

	int leave = (n - t - d) * (m - l - r) - k;

	if(leave < 0) {
		cout << 0 << "\n";
		return;
	} 

	for(int i = 1; i < 2 * n; ++i){
		for(int j = 1; j < 2 * m; ++j){
			pre[i][j] += pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
		}
	}

	int ans = 0;
	int a = n - l - r, b = m - t - d;

	//cout << l << " " << r << " " << t << " " << d << " " << leave << "\n";

	for(int i = n - d; i <= 2 * n - 1 - d; ++i){
		for(int j = m - r; j <= 2 * m - 1 - r; ++j){
			if(pre[i][j] - pre[i - a][j] - pre[i][j - b] + pre[i - a][j - b] == leave) ++ans;
		}
	}

	cout << ans << "\n";
	return;
}

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

	int t = 1;
	cin >> t;
	while(t--) solve();

	return 0;
}