https://atcoder.jp/contests/abc054/tasks/abc054_d
// https://atcoder.jp/contests/abc054/tasks/abc054_d
// 背包
// 这里开始的时候数据规模想错了, 所以用了map, 实际上可以用数组 (40 * 10)^2 * 40
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 2e9;
void solv()
{
map<PII, int> f[2];
f[0][{0, 0}] = 0;
int n, ma, mb;
cin >> n >> ma >> mb;
for (int ii = 1; ii <= n; ii ++)
{
int a, b, c;
int i = ii&1;
cin >> a >> b >> c;
for (auto const &it: f[i^1])
{
PII ab = it.first;
PII ab2 = {ab.first + a, ab.second + b};
if (!f[i].count(ab)) f[i][ab] = it.second;
else f[i][ab] = min(f[i][ab], it.second);
if (!f[i].count(ab2)) f[i][ab2] = it.second + c;
else f[i][ab2] = min(f[i][ab2], it.second + c);
}
f[i^1].clear();
}
int ans = INF;
for (auto const &it: f[n&1])
{
int a = it.first.first, b = it.first.second, c = it.second;
if (a && b && ma*b==mb*a) ans = min(ans, c);
}
cout << (ans==INF ? -1: ans) << endl;
}
int main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T --)
{
solv();
}
return 0;
}