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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner

/**
 * タイプAの物質とタイプBの物質の混合比が Ma:Mb
 * sum(a_)*mb = sum(b_)*ma
 * よって、a * mb - b * maの合計が0になる組み合わせを探す。
 * cとnについての動的計画法で行う。
 *
 * 発散するかに見えるが、
 * -100 < a * mb - b * ma < 100
 * costの上限は、N * c <= 40 *100
 * 分量の状態は200*40以下
 * で間に合いそう。
 */
fun main(args: Array<String>) {
    class Drug(a: Int, b: Int, val cost: Int, ma: Int, mb: Int) {       // 1≦ai,bi≦10, 1≦ci≦100
        val excess = a * mb - b * ma
    }

    val (n, drugs) = Scanner(System.`in`).use {
        val (n, ma, mb) = Triple(it.nextInt(), it.nextInt(), it.nextInt())    // 1≦N≦40  1≦Ma,Mb≦10
        val drugs: List<Drug> = Array(n) {n ->
            Drug(it.nextInt(), it.nextInt(), it.nextInt(), ma, mb)
        }.toList()

        n to drugs
    }

    val maximumCost = drugs.sumBy { it.cost }// + 1      // N * c <= 40 *100
    // 初期条件: memo[0] = setOf(0)
    val zeroSet = emptySet<Int>()
    val initialMemo: List<Set<Int>> = Array(maximumCost, {zeroSet}).toList()//.apply{ this[0] = setOf(0) }.toList()
    // 漸化式: cell[i+1][c] = cell[i][c] + cell[i][c - drugs[i+1].cost].map{ + drugs[i+1].excess}
    val excessMemo = (0 until n).fold(initialMemo) { memo, i ->
        // cost == 0 生成量0で目的を満たさないので判定にかけない。
        (0 until memo.size).asSequence().map {cost ->   // 先頭を除いたので、cost は実際のコストより1少ない。
            when {
                cost - drugs[i].cost < -1 -> memo[cost]
                cost - drugs[i].cost == -1 -> memo[cost] + setOf(drugs[i].excess)
            // 新しいオブジェクトが生成されるので、変更されたnextMemoの要素をを誤ってmemoの要素として参照することはない。
                else -> memo[cost] + memo[cost - drugs[i].cost].map{ it + drugs[i].excess }
            }
        }.takeWhile { 0 !in it }.toList()    // 条件を満たすなら、それ以上のcostについては考慮する必要がない。
        // これには条件を満たすcost自身が含まれない。
    }

    //条件を満たすものが含まれない分、sizeはlastIndexより1大きいので正解になる。
    println(if (excessMemo.size == maximumCost) -1 else excessMemo.size + 1)    // 先頭を除いた分を足す。
}