#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
#define fr(i,b,e) for(int i=b; i<e; ++i)
#define f(i,e) fr(i,0,e)
#define loop(i,e) for(int i=0; i<e; ++i)
#define ll long long
#define ull unsigned long long
#define str string
void setio(){
    ios::sync_with_stdio(false); 
    cin.tie(NULL);
}
int main(){
    setio();
    int t;
    cin>>t;
    while(t--){
       int x,y;
       str s;
       cin>>x>>y>>s; 
       int n=s.length();
       int u=0,r=0,d=0,l=0;
       bool f1=0,f2=0;
       f(i,n){
           if(s[i]=='R'){
               r++;
           }
           else if(s[i]=='L'){
               l--;
           }
           else if(s[i]=='U'){
               u++;
           }
           else{
               d--;
           }
       }
       if(x>=0 && x<=r){
           f1=1;
       }
       if(x<0 && x>=l){
           f1=1;
       }
       if(y>=0 && y<=u){
           f2=1;
       }
       if(y<0 && y>=d){
           f2=1;
       }
       if(f1 && f2){
           cout<<"YES"<<endl;
       }
       else{
           cout<<"NO"<<endl;
       }
    }
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
#define fr(i,b,e) for(int i=b; i<e; ++i)
#define f(i,e) fr(i,0,e)
#define loop(i,e) for(int i=0; i<e; ++i)
#define ll long long
#define ull unsigned long long
#define str string
void setio(){
    ios::sync_with_stdio(false); 
    cin.tie(NULL);
}
int main(){
    setio();
    int t;
    cin>>t;
    while(t--){
       int x,y;
       str s;
       cin>>x>>y>>s; 
       int n=s.length();
       int u=0,r=0,d=0,l=0;
       bool f1=0,f2=0;
       f(i,n){
           if(s[i]=='R'){
               r++;
           }
           else if(s[i]=='L'){
               l--;
           }
           else if(s[i]=='U'){
               u++;
           }
           else{
               d--;
           }
       }
       if(x>=0 && x<=r){
           f1=1;
       }
       if(x<0 && x>=l){
           f1=1;
       }
       if(y>=0 && y<=u){
           f2=1;
       }
       if(y<0 && y>=d){
           f2=1;
       }
       if(f1 && f2){
           cout<<"YES"<<endl;
       }
       else{
           cout<<"NO"<<endl;
       }
    }
    return 0;
}

```