#include <bits/stdc++.h>
using namespace std;

const int nax = 2e5 + 5;
const int pax = 1e5 + 5;
const int INF = 2e9;
using ll = long long;
using pd = pair<long double, long double>;
using pi = pair<int, int>;
using pll = pair<ll, ll>;
const long double eps = 1e-7;
const int MOD = 1e9 + 7;

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t, px, py;
	string s;
	cin >> t;
	//unordered_map<char, pi> dir = { {'L', {-1, 0}}, {'R', {1, 0}}, {'U', {0, 1}}, {'D', {0, -1}} };
	while (t--) {
		int mxx = 0, mnx = 0, mny = 0, mxy = 0;
		//int cx = 0, cy = 0;
		cin >> px >> py >> s;
		for (int i = 0; i < s.length(); i++) {
			if (s[i] == 'L') {
				mnx--;
			}
			else if (s[i] == 'R') {
				mxx++;
			}
			else if (s[i] == 'U') {
				mxy++;
			}
			else {
				mny--;
			}
		}
		if ((px >= mnx) && (px <= mxx) && (py >= mny) && (py <= mxy)) {
			cout << "YES\n";
		}
		else cout << "NO\n";
	}
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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 <bits/stdc++.h>
using namespace std;

const int nax = 2e5 + 5;
const int pax = 1e5 + 5;
const int INF = 2e9;
using ll = long long;
using pd = pair<long double, long double>;
using pi = pair<int, int>;
using pll = pair<ll, ll>;
const long double eps = 1e-7;
const int MOD = 1e9 + 7;

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t, px, py;
	string s;
	cin >> t;
	//unordered_map<char, pi> dir = { {'L', {-1, 0}}, {'R', {1, 0}}, {'U', {0, 1}}, {'D', {0, -1}} };
	while (t--) {
		int mxx = 0, mnx = 0, mny = 0, mxy = 0;
		//int cx = 0, cy = 0;
		cin >> px >> py >> s;
		for (int i = 0; i < s.length(); i++) {
			if (s[i] == 'L') {
				mnx--;
			}
			else if (s[i] == 'R') {
				mxx++;
			}
			else if (s[i] == 'U') {
				mxy++;
			}
			else {
				mny--;
			}
		}
		if ((px >= mnx) && (px <= mxx) && (py >= mny) && (py <= mxy)) {
			cout << "YES\n";
		}
		else cout << "NO\n";
	}
	return 0;
}

```