The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.

Koa the Koala is at the beach!

The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.

She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.

Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:

  * For a total of k seconds, each second, tide increases all depths by 1.

  * Then, for a total of k seconds, each second, tide decreases all depths by 1.

  * This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).

Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding. 




At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:

  * In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.

  * As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.

  * Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.

Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there). 




Koa wants to know whether she can go from the shore to the island. Help her!

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.

The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.

The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.

It is guaranteed that the sum of n over all test cases does not exceed 100.

Output

For each test case:

Print Yes if Koa can get from the shore to the island, and No otherwise.

You may print each letter in any case (upper or lower).

Example

Input


7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2


Output


Yes
No
Yes
Yes
Yes
No
No

Note

In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.

In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].

Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:

  * Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i]. 
  * At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1. 
  * At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1. 
  * At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1. 
  * At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4. 
  * At t = 4 Koa is at x = 3 and she made it! 



We can show that in test case 2 Koa can't get to the island.
import java.io.*;
import java.util.*;
import java.math.*;

 
 public class Main	{
	static class node{
		int a, b;
		public node(int x, int y) {
			a = x;
			b = y;
		}
	}
   static class Pair implements Comparable<Pair>{

	   int a;
	   int b;
//	   int ind;
//	   public Pair(int x, long y) {a = x;b=y;}
	   public Pair(int x, int y) {a = x;b=y;}
//	   public Pair(int x,int y, int z){a=x;b=y;ind = z;}
	   public int compareTo(Pair p){
		   return b - p.b;
	   }
//		@Override
//		public int hashCode() {
//			final int prime = 31;
//			int result = 1;
//			result = prime * result + a;
//			result = prime * result + b;
//		   
//			return result;
//		}
//		@Override
//		public boolean equals(Object obj) {
//			Pair cur = (Pair)obj;
//			if((a==cur.a && b==cur.b))return true;
//			return false;
//		}
 }
  static class cell{
	  int a, b;
	  public cell(int x, int y) {
		  a = x;
		  b = y;
	  }
  }
static class TrieNode{
	TrieNode left, right;
	int cnt;
	public TrieNode() {
		cnt = 0;
	}
}
  public static long gcd(long a,long b)
  {
    if(a<b)
      return gcd(b,a);
    if(b==0)
      return a;
    return gcd(b,a%b);
    
  }
  static long lcm(int a,int b) {
	  return  a*b / gcd(a,b);
  }
  
   
    public static void main(String[] args) throws Exception {
       new Thread(null, null, "Anshum Gupta", 99999999) {
            public void run() {
                try {
                    solve();
                } catch(Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();
    }

static long pow(long x,long y){
    if(y == 0)return 1;
    if(y==1)return x;
    long a = pow(x,y/2);
    a = (a*a)%mod;
    if(y%2==0){
        return a;
    }
    return (a*x)%mod;
}

static long[]fact,inv_fact;
static long my_inv(long a) {
	return pow(a,mod-2);
}
static long bin(int a,int b) {
	if(a < b || a<0 || b<0)return 0;
    return ((fact[a]*inv_fact[a-b])%mod * inv_fact[b])%mod;
}
	
static void make_facts() {
  fact=new long[mxN];
  inv_fact = new long[mxN];
  fact[0]=inv_fact[0]=1L;
  for(int i=1;i<mxN;i++) {
	   fact[i] = (i*fact[i-1])%mod;
	   inv_fact[i] = my_inv(fact[i]);
  }
}
static void lazy(int st, int e, int tn) {
	if(lazy[tn] != INF) {
		tree[tn] = Math.min(tree[tn], lazy[tn]);
		if(st != e) {
			lazy[2*tn] = Math.min(lazy[2*tn], lazy[tn]);
			lazy[2*tn+1] = Math.min(lazy[2*tn+1], lazy[tn]);
		}
		lazy[tn] = INF;
	}
}
static void update(int ind, int val, int st, int e, int tn) {
	if(st == e) {
		tree[tn] = val;
		return;
	}
	int mid = (st + e) >> 1;	
  	if(ind <= mid)
		update(ind, val, st, mid, 2*tn);
  	else
		update(ind, val, mid+1, e, 2*tn+1);
	tree[tn] = Math.min(tree[2*tn], tree[2*tn+1]);
}
static int query(int l, int r, int st, int e, int tn) {
	if(st > r || e < l)return INF;
	
	if(st >= l && e <= r) {
		return tree[tn];
	}

	int mid = (st + e) >> 1;
	int x = query(l, r, st, mid, 2 * tn);
	int y = query(l, r, mid+1, e, 2 * tn + 1);
	return Math.min(x, y);
}
static final long mxx = (long)(1e18+5);
static final int mxN = (int)(1e5);
static final int mxV = (int)(1e5+5);
static final long mod = 998244353;//(long)(1e9+7); //
static int[]tree, lazy;
static final int INF = (int)1e9+5;
static ArrayList<ArrayList<Integer>> adj;
static boolean[] dp2;
static int n, m, b, c, q, t, k, l;
static int[]d;
static int[][]cnt;
static boolean[]vis;
	public static void solve() throws Exception {
   // solve the problem here
   MyScanner s = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out), true);
        int tc = s.nextInt();
        while(tc-->0){
        	n = s.nextInt();
        	k = s.nextInt();
        	l = s.nextInt();
        	d = new int[n+1];
        	ArrayList<Integer> safe = new ArrayList<Integer>();
        	safe.add(0);
        	for(int i=1; i<=n; i++) {
        		d[i] = s.nextInt();
        		if(d[i] + k <= l) {
        			safe.add(i);
        		}
        	}
        	safe.add(n+1);
        	boolean ok = true;
        	for(int i=1; i<safe.size() && ok; i++) {
        		int tide = k;
        		boolean down = true;
        		for(int j = safe.get(i-1) + 1; j < safe.get(i); j++) {
        			tide += down ? -1 : +1;
        			if(down) 
        				tide -= Math.max(0, d[j] + tide - l);
        			if(tide < 0 || d[j] + tide > l) {
        				ok = false;
        				break;
        			}
        			if(tide == 0)
        				down = false;
        		}
        	}
        	out.println(ok ? "yes" : "no");
        }
        	
           
        out.flush();
}
 
     
 
 
 
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
 
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }
         
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------
}
