Problem:
Score : 400 points

Problem Statement

Dolphin is planning to generate a small amount of a certain chemical substance C.

In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.

He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.

The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.

The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).

Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.

Find the minimum amount of money required to generate the substance C.

If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.

Constraints

1≦N≦40

1≦a_i,b_i≦10

1≦c_i≦100

1≦M_a,M_b≦10

gcd(M_a,M_b)=1

a_i, b_i, c_i, M_a and M_b are integers.

Input

The input is given from Standard Input in the following format:

N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N

Output

Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print -1 instead.

Sample Input 1

3 1 1
1 2 1
2 1 2
3 3 10

Sample Output 1

3

The amount of money spent will be minimized by purchasing the packages of chemicals 1 and 2.

In this case, the mixture of the purchased chemicals will contain 3 grams of the substance A and 3 grams of the substance B, which are in the desired ratio: 3:3=1:1.

The total price of these packages is 3 yen.

Sample Input 2

1 1 10
10 10 10

Sample Output 2

-1

The ratio 1:10 of the two substances A and B cannot be satisfied by purchasing any combination of the packages. Thus, the output should be -1.

Solution in Scala (2.11.7) (judge verdict: Accepted)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    // タイプAとタイプBの物質の混合比がM_a : M_b
    // A の重さが a の時, Bの重さ b は b = (a * M_b) / M_a { (a * M_b) % M_a == 0 } である必要がある.

    // 最初, dp(i)(a)(b) = Inf としておくと, その最小価格が存在しない時には Infである.
    // 条件を満たす全ての dpが Infなら, 作れないということである。
    // i個の薬品を使って, a, b の重さを作る時の 価格は dp(i)(a)(b) で表せるとすると,
    // i番目の薬品を加える時と加えない時の２種類で更新できる。
    // dp(i)(a)(b) = dp(i-1)(a)(b) // 加えない場合
    // dp(i)(a+A(i))(b+B(i)) = dp(i-1)(a)(b) + C(i)
    val INF = Int.MaxValue / 10
    val N, Ma, Mb = sc.nextInt()
    val amax = 10
    val bmax = 10
    val nmax = 40
    val dp = Array.fill(N+1, nmax*amax+1, nmax*bmax+1)(INF)

    dp(0)(0)(0) = 0

    val A, B, C = new Array[Int](N)
    (0 until N) foreach { i =>
      A(i) = sc.nextInt()
      B(i) = sc.nextInt()
      C(i) = sc.nextInt()
    }

    (0 until N) foreach { i =>
      (0 to nmax*amax - A(i)) foreach { a =>
        (0 to nmax*bmax - B(i)) foreach { b =>
          dp(i+1)(a)(b) = Math.min(dp(i+1)(a)(b), dp(i)(a)(b))
          dp(i+1)(a + A(i))(b + B(i)) = Math.min(dp(i+1)(a + A(i))(b + B(i)), dp(i)(a)(b) + C(i))
        }
      }
    }

    var min = INF
    (1 until nmax*amax) foreach { a =>
      (1 until nmax*bmax) foreach { b =>
        if (a * Mb == b * Ma)
          min = Math.min(min, dp(N)(a)(b))
      }
    }

    println(if (min == INF) -1
    else min)
  }
}