#include <array>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <functional>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
using namespace std;
using ll = long long; using pii = pair<int, int>; using pll = pair<ll, ll>; using vpii = vector<pii>; using vpll = vector<pll>;
template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { for (const auto &x : v) os << x << " "; return os; }

void solve() {
	int px, py;
	cin >> px >> py;
	string s;
	cin >> s;
	int xmn = 0, xmx = 0, ymn = 0, ymx = 0;
	for (char c : s) {
		if (c=='U')
			ymx++;
		else if (c=='D')
			ymn--;
		else if (c=='R')
			xmx++;
		else
			xmn--;
	}
	bool ans = xmn <= px && px <= xmx && ymn <= py && py <= ymx;
	cout << (ans ? "YES" : "NO") << endl;
}

int main(int argc, char const *argv[]) {
	int T;
	cin >> T;
	for (int t=0; t<T; t++)
		solve();
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You were dreaming that you are traveling to a planet named Planetforces on your personal spaceship. Unfortunately, its piloting system was corrupted and now you need to fix it in order to reach Planetforces.

<image>

Space can be represented as the XY plane. You are starting at point (0, 0), and Planetforces is located in point (p_x, p_y).

The piloting system of your spaceship follows its list of orders which can be represented as a string s. The system reads s from left to right. Suppose you are at point (x, y) and current order is s_i: 

  * if s_i = U, you move to (x, y + 1); 
  * if s_i = D, you move to (x, y - 1); 
  * if s_i = R, you move to (x + 1, y); 
  * if s_i = L, you move to (x - 1, y). 



Since string s could be corrupted, there is a possibility that you won't reach Planetforces in the end. Fortunately, you can delete some orders from s but you can't change their positions.

Can you delete several orders (possibly, zero) from s in such a way, that you'll reach Planetforces after the system processes all orders?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of two lines. The first line in each test case contains two integers p_x and p_y (-10^5 ≤ p_x, p_y ≤ 10^5; (p_x, p_y) ≠ (0, 0)) — the coordinates of Planetforces (p_x, p_y).

The second line contains the string s (1 ≤ |s| ≤ 10^5: |s| is the length of string s) — the list of orders.

It is guaranteed that the sum of |s| over all test cases does not exceed 10^5.

Output

For each test case, print "YES" if you can delete several orders (possibly, zero) from s in such a way, that you'll reach Planetforces. Otherwise, print "NO". You can print each letter in any case (upper or lower).

Example

Input


6
10 5
RRRRRRRRRRUUUUU
1 1
UDDDRLLL
-3 -5
LDLDLDDDR
1 2
LLLLUU
3 -2
RDULRLLDR
-1 6
RUDURUUUUR


Output


YES
YES
YES
NO
YES
NO

Note

In the first case, you don't need to modify s, since the given s will bring you to Planetforces.

In the second case, you can delete orders s_2, s_3, s_4, s_6, s_7 and s_8, so s becomes equal to "UR".

In the third test case, you have to delete order s_9, otherwise, you won't finish in the position of Planetforces.

### Response

```cpp
#include <array>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <functional>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
using namespace std;
using ll = long long; using pii = pair<int, int>; using pll = pair<ll, ll>; using vpii = vector<pii>; using vpll = vector<pll>;
template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { for (const auto &x : v) os << x << " "; return os; }

void solve() {
	int px, py;
	cin >> px >> py;
	string s;
	cin >> s;
	int xmn = 0, xmx = 0, ymn = 0, ymx = 0;
	for (char c : s) {
		if (c=='U')
			ymx++;
		else if (c=='D')
			ymn--;
		else if (c=='R')
			xmx++;
		else
			xmn--;
	}
	bool ans = xmn <= px && px <= xmx && ymn <= py && py <= ymx;
	cout << (ans ? "YES" : "NO") << endl;
}

int main(int argc, char const *argv[]) {
	int T;
	cin >> T;
	for (int t=0; t<T; t++)
		solve();
	return 0;
}

```