Problem: On the way home, Karen decided to stop by the supermarket to buy some groceries.

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She needs to buy a lot of goods, but since she is a student her budget is still quite limited. In fact, she can only spend up to b dollars.

The supermarket sells n goods. The i-th good can be bought for ci dollars. Of course, each good can only be bought once.

Lately, the supermarket has been trying to increase its business. Karen, being a loyal customer, was given n coupons. If Karen purchases the i-th good, she can use the i-th coupon to decrease its price by di. Of course, a coupon cannot be used without buying the corresponding good.

There is, however, a constraint with the coupons. For all i ≥ 2, in order to use the i-th coupon, Karen must also use the xi-th coupon (which may mean using even more coupons to satisfy the requirement for that coupon).

Karen wants to know the following. What is the maximum number of goods she can buy, without exceeding her budget b?

Input

The first line of input contains two integers n and b (1 ≤ n ≤ 5000, 1 ≤ b ≤ 109), the number of goods in the store and the amount of money Karen has, respectively.

The next n lines describe the items. Specifically:

  * The i-th line among these starts with two integers, ci and di (1 ≤ di < ci ≤ 109), the price of the i-th good and the discount when using the coupon for the i-th good, respectively. 
  * If i ≥ 2, this is followed by another integer, xi (1 ≤ xi < i), denoting that the xi-th coupon must also be used before this coupon can be used. 

Output

Output a single integer on a line by itself, the number of different goods Karen can buy, without exceeding her budget.

Examples

Input

6 16
10 9
10 5 1
12 2 1
20 18 3
10 2 3
2 1 5


Output

4


Input

5 10
3 1
3 1 1
3 1 2
3 1 3
3 1 4


Output

5

Note

In the first test case, Karen can purchase the following 4 items:

  * Use the first coupon to buy the first item for 10 - 9 = 1 dollar. 
  * Use the third coupon to buy the third item for 12 - 2 = 10 dollars. 
  * Use the fourth coupon to buy the fourth item for 20 - 18 = 2 dollars. 
  * Buy the sixth item for 2 dollars. 



The total cost of these goods is 15, which falls within her budget. Note, for example, that she cannot use the coupon on the sixth item, because then she should have also used the fifth coupon to buy the fifth item, which she did not do here.

In the second test case, Karen has enough money to use all the coupons and purchase everything.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			long startTime = System.currentTimeMillis();
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			FastReader in = new FastReader(inputStream);
			Output out = new Output(outputStream);
			CKarenAndSupermarket solver = new CKarenAndSupermarket();
			solver.solve(1, in, out);
			out.close();
			System.err.println(System.currentTimeMillis()-startTime+"ms");
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<28);
		thread.start();
		thread.join();
	}

	static class CKarenAndSupermarket {
		private final long linf = 4_000_000_000_000_000_000L;
		int n;
		int[] size;
		long[] c;
		long[] d;
		long[][] f;
		long[][] g;
		ArrayList<Integer>[] graph;

		public CKarenAndSupermarket() {
		}

		public void dp(int u) {
			size[u] = 1;
			for(int v: graph[u]) {
				dp(v);
				size[u] += size[v];
			}
			int m = graph[u].size(), csize = 1;
			long[][] ftmp = new long[2][size[u]+1];
			long[][] gtmp = new long[2][size[u]+1];
			Arrays.fill(ftmp[0], linf);
			Arrays.fill(gtmp[0], linf);
			ftmp[0][1] = c[u]-d[u];
			gtmp[0][0] = 0;
			gtmp[0][1] = c[u];
			int cur = 1, prev = 0;
			for(int i = 0; i<m; i++, cur = prev, prev ^= 1) {
				int v = graph[u].get(i);
				Arrays.fill(ftmp[cur], linf);
				Arrays.fill(gtmp[cur], linf);
				for(int j = 0; j<=csize; j++) {
					for(int k = 0; k<=size[v]; k++) {
						ftmp[cur][j+k] = Math.min(ftmp[cur][j+k], ftmp[prev][j]+Math.min(f[v][k], g[v][k]));
						gtmp[cur][j+k] = Math.min(gtmp[cur][j+k], gtmp[prev][j]+g[v][k]);
					}
				}
				csize += size[v];
			}
			for(int i = 0; i<=csize; i++) {
				f[u][i] = ftmp[prev][i];
				g[u][i] = gtmp[prev][i];
			}
			f[u][0] = 0;
		}

		public void solve(int kase, InputReader in, Output pw) {
			n = in.nextInt();
			long b = in.nextInt();
			c = new long[n];
			d = new long[n];
			graph = new ArrayList[n];
			for(int i = 0; i<n; i++) {
				graph[i] = new ArrayList<>();
			}
			for(int i = 0; i<n; i++) {
				c[i] = in.nextInt();
				d[i] = in.nextInt();
				if(i>0) {
					graph[in.nextInt()-1].add(i);
				}
			}
			size = new int[n];
			f = new long[n][n+1];
			g = new long[n][n+1];
			for(int i = 0; i<n; i++) {
				Arrays.fill(f[i], linf);
				Arrays.fill(g[i], linf);
			}
			dp(0);
//			Utilities.Debug.dbg(f);
//			Utilities.Debug.dbg(g);
			for(int i = 0; i<=n; i++) {
				if(Math.min(f[0][i], g[0][i])>b) {
					pw.println(i-1);
					return;
				}
			}
			pw.println(n);
		}

	}

	interface InputReader {
		int nextInt();

	}

	static class FastReader implements InputReader {
		final private int BUFFER_SIZE = 1<<16;
		private final DataInputStream din;
		private final byte[] buffer;
		private int bufferPointer;
		private int bytesRead;

		public FastReader(InputStream is) {
			din = new DataInputStream(is);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public int nextInt() {
			int ret = 0;
			byte c = skipToDigit();
			boolean neg = (c=='-');
			if(neg) {
				c = read();
			}
			do {
				ret = ret*10+c-'0';
			} while((c = read())>='0'&&c<='9');
			if(neg) {
				return -ret;
			}
			return ret;
		}

		private boolean isDigit(byte b) {
			return b>='0'&&b<='9';
		}

		private byte skipToDigit() {
			byte ret;
			while(!isDigit(ret = read())&&ret!='-') ;
			return ret;
		}

		private void fillBuffer() {
			try {
				bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			}catch(IOException e) {
				e.printStackTrace();
				throw new InputMismatchException();
			}
			if(bytesRead==-1) {
				buffer[0] = -1;
			}
		}

		private byte read() {
			if(bytesRead==-1) {
				throw new InputMismatchException();
			}else if(bufferPointer==bytesRead) {
				fillBuffer();
			}
			return buffer[bufferPointer++];
		}

	}

	static class Output implements Closeable, Flushable {
		public StringBuilder sb;
		public OutputStream os;
		public int BUFFER_SIZE;
		public String lineSeparator;

		public Output(OutputStream os) {
			this(os, 1<<16);
		}

		public Output(OutputStream os, int bs) {
			BUFFER_SIZE = bs;
			sb = new StringBuilder(BUFFER_SIZE);
			this.os = new BufferedOutputStream(os, 1<<17);
			lineSeparator = System.lineSeparator();
		}

		public void println(int i) {
			println(String.valueOf(i));
		}

		public void println(String s) {
			sb.append(s);
			println();
		}

		public void println() {
			sb.append(lineSeparator);
		}

		private void flushToBuffer() {
			try {
				os.write(sb.toString().getBytes());
			}catch(IOException e) {
				e.printStackTrace();
			}
			sb = new StringBuilder(BUFFER_SIZE);
		}

		public void flush() {
			try {
				flushToBuffer();
				os.flush();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

		public void close() {
			flush();
			try {
				os.close();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

	}

	static class Utilities {
		public static class Debug {
			public static final boolean LOCAL = System.getProperty("ONLINE_JUDGE")==null;

			private static <T> String ts(T t) {
				if(t==null) {
					return "null";
				}
				try {
					return ts((Iterable) t);
				}catch(ClassCastException e) {
					if(t instanceof int[]) {
						String s = Arrays.toString((int[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof long[]) {
						String s = Arrays.toString((long[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof char[]) {
						String s = Arrays.toString((char[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof double[]) {
						String s = Arrays.toString((double[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof boolean[]) {
						String s = Arrays.toString((boolean[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}
					try {
						return ts((Object[]) t);
					}catch(ClassCastException e1) {
						return t.toString();
					}
				}
			}

			private static <T> String ts(T[] arr) {
				StringBuilder ret = new StringBuilder();
				ret.append("{");
				boolean first = true;
				for(T t: arr) {
					if(!first) {
						ret.append(", ");
					}
					first = false;
					ret.append(ts(t));
				}
				ret.append("}");
				return ret.toString();
			}

			private static <T> String ts(Iterable<T> iter) {
				StringBuilder ret = new StringBuilder();
				ret.append("{");
				boolean first = true;
				for(T t: iter) {
					if(!first) {
						ret.append(", ");
					}
					first = false;
					ret.append(ts(t));
				}
				ret.append("}");
				return ret.toString();
			}

			public static void dbg(Object... o) {
				if(LOCAL) {
					System.err.print("Line #"+Thread.currentThread().getStackTrace()[2].getLineNumber()+": [");
					for(int i = 0; i<o.length; i++) {
						if(i!=0) {
							System.err.print(", ");
						}
						System.err.print(ts(o[i]));
					}
					System.err.println("]");
				}
			}

		}

	}
}