Problem:
Score : 800 points

Problem Statement

Takahashi and Aoki will take N exams numbered 1 to N.
They have decided to compete in these exams.
The winner will be determined as follows:

For each exam i, Takahashi decides its importance c_i, which must be an integer between l_i and u_i (inclusive).

Let A be \sum_{i=1}^{N} c_i \times (Takahashi's score on Exam i), and B be \sum_{i=1}^{N} c_i \times (Aoki's score on Exam i). Takahashi wins if A \geq B, and Aoki wins if A < B.

Takahashi knows that Aoki will score b_i on Exam i, with his supernatural power.

Takahashi himself, on the other hand, will score 0 on all the exams without studying more. For each hour of study, he can increase his score on some exam by 1. (He can only study for an integer number of hours.)
However, he cannot score more than X on an exam, since the perfect score for all the exams is X.

Print the minimum number of study hours required for Takahashi to win.

Constraints

1 \leq N \leq 10^5

1 \leq X \leq 10^5

0 \leq b_i \leq X (1 \leq i \leq N)

1 \leq l_i \leq u_i \leq 10^5 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X
b_1 l_1 u_1
b_2 l_2 u_2
:
b_N l_N u_N

Output

Print the minimum number of study hours required for Takahashi to win.

Sample Input 1

2 100
85 2 3
60 1 1

Sample Output 1

115

One optimal strategy is as follows:

Choose c_1 = 3, c_2 = 1.

Study to score 100 on Exam 1 and 15 on Exam 2.

Then, A = 3 \times 100 + 1 \times 15 = 315, B = 3 \times 85 + 1 \times 60 = 315 and Takahashi will win.

Sample Input 2

2 100
85 2 3
60 10 10

Sample Output 2

77

Sample Input 3

1 100000
31415 2718 2818

Sample Output 3

31415

Sample Input 4

10 1000
451 4593 6263
324 310 6991
378 1431 7068
71 1757 9218
204 3676 4328
840 6221 9080
684 1545 8511
709 5467 8674
862 6504 9835
283 4965 9980

Sample Output 4

2540

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Diagnostics;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using System.Numerics;
using static System.Math;
using Number = System.Int32;

namespace Program {
	public class Solver {
		Random rnd = new Random();
		public void Solve() {
			var n = ri;
			var X = rl;
			var a = Enumerate(n, x => new long[] { rl, rl, rl });
			var sum = 0L;
			foreach (var x in a)
				sum += x[0] * x[1];
			if (sum == 0) {
				Console.WriteLine(0);
				return;
			}
			Debug.WriteLine(sum);
			Array.Sort(a, (l, r) =>
				(l[2] * (X - l[0]) + l[0] * l[1]).CompareTo(r[2] * (X - r[0]) + r[0] * r[1])
 			);
			Array.Reverse(a);
			var b = new long[n + 1];
			for (int i = 0; i < n; i++)
				b[i + 1] = b[i] + a[i][2] * (X - a[i][0]) + a[i][0] * a[i][1];
			Debug.WriteLine(b.AsJoinedString());
			var min = INF;
			var k = n - 1;
			for (int _ = 0; _ < 2; _++) {
				for (int i = 0; i < n; i++) {
					while (true) {
						var val = b[k];
						var p = a[i][0];
						var L = a[i][1];
						var R = a[i][2];
						if (i < k) {
							val = b[k + 1];
							val -= R * (X - p) + p * L;
						}
						val -= sum;

						if (val >= 0) k--;
						else {
							val *= -1;
							if ((val + L - 1) / L <= p) {
								min = Min(min, k * X + (val + L - 1) / L);
							}
							else {
								val -= L * p;
								if ((val + R - 1) / R <= X - p)
									min = Min(min, k * X + p + (val + R - 1) / R);
							}
							break;
						}
					}
				}
			}
			Debug.WriteLine(k);
			Console.WriteLine(min);
		}
		const long INF = 1L << 60;
		static int[] dx = { -1, 0, 1, 0 };
		static int[] dy = { 0, 1, 0, -1 };
		int ri { get { return sc.Integer(); } }
		long rl { get { return sc.Long(); } }
		double rd { get { return sc.Double(); } }
		string rs { get { return sc.Scan(); } }
		public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

		static T[] Enumerate<T>(int n, Func<int, T> f) {
			var a = new T[n];
			for (int i = 0; i < a.Length; ++i) a[i] = f(i);
			return a;
		}
		static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
	}
}

#region main
static class Ex {
	static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
	static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
		return string.Join(st, ie);
	}
	static public void Main() {
		Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = true });
		var solver = new Program.Solver();
		/* 
		var t = new System.Threading.Thread(solver.Solve, 50000000);
		t.Start();
		t.Join();
		*/
		solver.Solve();
		Console.Out.Flush();
	}
}
#endregion
#region Ex
namespace Program.IO {
	using System.IO;
	using System.Text;
	using System.Globalization;

	public class Printer : StreamWriter {
		public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
		public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
	}

	public class StreamScanner {
		public StreamScanner(Stream stream) { str = stream; }

		public readonly Stream str;
		private readonly byte[] buf = new byte[1024];
		private int len, ptr;
		public bool isEof = false;
		public bool IsEndOfStream { get { return isEof; } }

		private byte read() {
			if (isEof) return 0;
			if (ptr >= len) {
				ptr = 0;
				if ((len = str.Read(buf, 0, 1024)) <= 0) {
					isEof = true;
					return 0;
				}
			}
			return buf[ptr++];
		}

		public char Char() {
			byte b = 0;
			do b = read(); while ((b < 33 || 126 < b) && !isEof);
			return (char)b;
		}
		public string Scan() {
			var sb = new StringBuilder();
			for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
			return sb.ToString();
		}
		public string ScanLine() {
			var sb = new StringBuilder();
			for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
			return sb.ToString();
		}
		public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
		public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
		public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
	}
}

#endregion


#region ModInt
/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt {
	/// <summary>
	/// 剰余を取る値．
	/// </summary>
	public const long Mod = (int)1e9 + 7;

	/// <summary>
	/// 実際の数値．
	/// </summary>
	public long num;
	/// <summary>
	/// 値が <paramref name="n"/> であるようなインスタンスを構築します．
	/// </summary>
	/// <param name="n">インスタンスが持つ値</param>
	/// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
	public ModInt(long n) { num = n; }
	/// <summary>
	/// このインスタンスの数値を文字列に変換します．
	/// </summary>
	/// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
	public override string ToString() { return num.ToString(); }
	public static ModInt operator +(ModInt l, ModInt r) { l.num += r.num; if (l.num >= Mod) l.num -= Mod; return l; }
	public static ModInt operator -(ModInt l, ModInt r) { l.num -= r.num; if (l.num < 0) l.num += Mod; return l; }
	public static ModInt operator *(ModInt l, ModInt r) { return new ModInt(l.num * r.num % Mod); }
	public static implicit operator ModInt(long n) { n %= Mod; if (n < 0) n += Mod; return new ModInt(n); }

	/// <summary>
	/// 与えられた 2 つの数値からべき剰余を計算します．
	/// </summary>
	/// <param name="v">べき乗の底</param>
	/// <param name="k">べき指数</param>
	/// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
	public static ModInt Pow(ModInt v, long k) { return Pow(v.num, k); }

	/// <summary>
	/// 与えられた 2 つの数値からべき剰余を計算します．
	/// </summary>
	/// <param name="v">べき乗の底</param>
	/// <param name="k">べき指数</param>
	/// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
	public static ModInt Pow(long v, long k) {
		long ret = 1;
		for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
			if ((k & 1) == 1) ret = ret * v % Mod;
		return new ModInt(ret);
	}
	/// <summary>
	/// 与えられた数の逆元を計算します．
	/// </summary>
	/// <param name="v">逆元を取る対象となる数</param>
	/// <returns>逆元となるような値</returns>
	/// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
	public static ModInt Inverse(ModInt v) { return Pow(v, Mod - 2); }
}
#endregion
#region Binomial Coefficient
public class BinomialCoefficient {
	public ModInt[] fact, ifact;
	public BinomialCoefficient(int n) {
		fact = new ModInt[n + 1];
		ifact = new ModInt[n + 1];
		fact[0] = 1;
		for (int i = 1; i <= n; i++)
			fact[i] = fact[i - 1] * i;
		ifact[n] = ModInt.Inverse(fact[n]);
		for (int i = n - 1; i >= 0; i--)
			ifact[i] = ifact[i + 1] * (i + 1);
		ifact[0] = ifact[1];
	}
	public ModInt this[int n, int r] {
		get {
			if (n < 0 || n >= fact.Length || r < 0 || r > n) return 0;
			return fact[n] * ifact[n - r] * ifact[r];
		}
	}
	public ModInt RepeatedCombination(int n, int k) {
		if (k == 0) return 1;
		return this[n + k - 1, k];
	}
}
#endregion