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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

/*

i-1番目まで調べて、物質aがAグラム、bがBグラムのときの最低価格をp[i][A][B]と置く。
A,Bは0から400まで、iは0から41まで

p[i][A][B] = 5000と置いておけば十分。

漸化式は、
p[i][A][B] = min(p[i-1][A-a[i-1]][B-b[i-1]] + c[i-1] , p[i-1][A][B])

p[0][0][0] = 0, 残りはp[0][A][B] = 5000。

*/
int a[40], b[40], c[40];

int p[41][401][401];
int memo[41][401][401];

int min(int i, int j){
	if(i<j){return i;}else{return j;}
}

int dp(int i,int A,int B){
	int PP;
	
	if(A<0){return 5000;}
	if(B<0){return 5000;}
	if(memo[i][A][B] == 1){
		return p[i][A][B];
	}
	PP = min(dp(i-1,A-a[i-1],B-b[i-1]) + c[i-1] , dp(i-1,A,B));
	memo[i][A][B] = 1;
	p[i][A][B] = PP;
	return PP;
}


int main() {
	int N,Ma,Mb,i,j,k,ans;
	scanf("%d %d %d",&N,&Ma,&Mb);
	for(i=0;i<N;i++){
		scanf("%d %d %d",&a[i],&b[i],&c[i]);
	}

	for(i=0;i<41;i++){
		for(j=0;j<401;j++){
			for(k=0;k<401;k++){
				memo[i][j][k] = 0;
				p[i][j][k] = 5000;
			}
		}
	}
	for(j=0;j<401;j++){
		for(k=0;k<401;k++){
			memo[0][j][k] = 1;
		}
	}
	p[0][0][0] = 0;

	ans = 5000;
	for(i=1;(i*Ma<=400) && (i*Mb<=400); i++){
		ans = min(ans, dp(N,Ma*i,Mb*i));
	}

	if(ans==5000){
		printf("-1\n");
	}else{
		printf("%d\n",ans);
	}
}