#include<bits/stdc++.h>
using namespace std;
#pragma region atcoder
//#include <atcoder/dsu>
//using namespace atcoder;
//using mint = modint998244353;
//using mint = modint1000000007;
#pragma endregion
#pragma region debug for var, v, vv
#define debug(var)  do{std::cerr << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cerr << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cerr << e << " "; } std::cerr << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){cerr << endl;int cnt = 0;for(const auto& v : vv){cerr << cnt << "th : "; view(v); cnt++;} cerr << endl;}
#pragma endregion
using ll = long long;
const ll mod = 1000000007;
const int inf = 1001001001;
const ll INF = 1001001001001001001ll; 
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
template<class T, class K>bool chmax(T &a, const K b) { if (a<b) { a=b; return 1; } return 0; }
template<class T, class K>bool chmin(T &a, const K b) { if (b<a) { a=b; return 1; } return 0; }
ll rudiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } //  20 / 3 == 7
ll rddiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // -20 / 3 == -7
ll power(ll a, ll p){ll ret = 1; while(p){if(p & 1){ret = ret * a;} a = a * a; p >>= 1;} return ret;}
ll modpow(ll a, ll p, ll m){ll ret = 1; while(p){if(p & 1){ret = ret * a % m;} a = a * a % m; p >>= 1;} return ret;}
/*---------------------------------------------------------------------------------------------------------------------------------*/
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    //cout << fixed << setprecision(20);
    int t; cin >> t;
    while(t--){
        int x, y; cin >> x >> y;
        string s; cin >> s;
        vector<int> cnt(4);
        int cx = 0, cy = 0;
        for(int i = 0; i < s.size(); i++){
            if(s[i] == 'R') cnt[0]++, cx++;
            if(s[i] == 'L') cnt[1]++, cx--;
            if(s[i] == 'U') cnt[2]++, cy++;
            if(s[i] == 'D') cnt[3]++, cy--;
        }
        if(cx < x && cnt[1] + cx >= x) cx = x;
        if(cx > x && cx - cnt[0] <= x) cx = x;
        if(cy < y && cy + cnt[3] >= y) cy = y;
        if(cy > y && cy - cnt[2] <= y) cy = y;
        cout << ((cx == x && cy == y) ? "YES" : "NO") << endl;
    }
}       
/*
   * review you code when you get WA (typo? index?)
   * int overflow, array bounds
   * special cases (n=1?)
*/
### Prompt

Please formulate 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;
#pragma region atcoder
//#include <atcoder/dsu>
//using namespace atcoder;
//using mint = modint998244353;
//using mint = modint1000000007;
#pragma endregion
#pragma region debug for var, v, vv
#define debug(var)  do{std::cerr << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cerr << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cerr << e << " "; } std::cerr << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){cerr << endl;int cnt = 0;for(const auto& v : vv){cerr << cnt << "th : "; view(v); cnt++;} cerr << endl;}
#pragma endregion
using ll = long long;
const ll mod = 1000000007;
const int inf = 1001001001;
const ll INF = 1001001001001001001ll; 
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
template<class T, class K>bool chmax(T &a, const K b) { if (a<b) { a=b; return 1; } return 0; }
template<class T, class K>bool chmin(T &a, const K b) { if (b<a) { a=b; return 1; } return 0; }
ll rudiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } //  20 / 3 == 7
ll rddiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // -20 / 3 == -7
ll power(ll a, ll p){ll ret = 1; while(p){if(p & 1){ret = ret * a;} a = a * a; p >>= 1;} return ret;}
ll modpow(ll a, ll p, ll m){ll ret = 1; while(p){if(p & 1){ret = ret * a % m;} a = a * a % m; p >>= 1;} return ret;}
/*---------------------------------------------------------------------------------------------------------------------------------*/
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    //cout << fixed << setprecision(20);
    int t; cin >> t;
    while(t--){
        int x, y; cin >> x >> y;
        string s; cin >> s;
        vector<int> cnt(4);
        int cx = 0, cy = 0;
        for(int i = 0; i < s.size(); i++){
            if(s[i] == 'R') cnt[0]++, cx++;
            if(s[i] == 'L') cnt[1]++, cx--;
            if(s[i] == 'U') cnt[2]++, cy++;
            if(s[i] == 'D') cnt[3]++, cy--;
        }
        if(cx < x && cnt[1] + cx >= x) cx = x;
        if(cx > x && cx - cnt[0] <= x) cx = x;
        if(cy < y && cy + cnt[3] >= y) cy = y;
        if(cy > y && cy - cnt[2] <= y) cy = y;
        cout << ((cx == x && cy == y) ? "YES" : "NO") << endl;
    }
}       
/*
   * review you code when you get WA (typo? index?)
   * int overflow, array bounds
   * special cases (n=1?)
*/
```