#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
typedef  long long ll;
double pi=3.1415926535898;
//double pi=acos(-1.0);
#define speed      ios::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
#define rng(i,a,b) for(int i=a;i>=int(b);i--)
#define rep(i,a,b) for(int i=a;i<=int(b);++i)
#define lop(i,n)   for(int i=0;i<int(n);i++)
#define ms(x,a)    memset(x,a,sizeof(x))
#define all(x)     (x).begin(),(x).end()
#define sz(x)      int(x.size())
#define pb         push_back
#define r0         return 0;
#define CC         continue;
#define bb(x)      cerr<< x <<endl;
#define pii        pair<int,int>
#define pis        pair<int,string>
#define vi         vector<int>
#define F          first
#define S          second
//#define max(a,b) a<b?b:a  __builtin_popcount;
//#define min(a,b) a<b?a:b
//freopen("output.txt","w",stdout),freopen("input.txt","r",stdin);
const ll M  = 1e9+7 ;
const int N = 50500;
const int inf = INT_MAX;
const double eps = 1e-6;
//////////////////////    Hi    ^_^  hack me if you can :P
int n,ma,mb;
int a[55];
int b[55];
int c[55];
int mem[41][401][401];
int dp (int i,int a_,int b_)
{
    if (i==n)
    {
        if (a_==0||b_==0)
            return 5000;
 
        int g = __gcd(a_,b_);
        if (a_/g==ma&&b_/g==mb)
        {
            return 0;
        }
        return 5000;
    }
    int &r=mem[i][a_][b_] ;
    if (r!=-1)
        return r;
    r = dp(i+1,a_,b_) ;
    r = min (r,dp(i+1,a_+a[i],b_+b[i])+c[i] );
    return r;
}
 
int main()
{
    cin>>n>>ma>>mb;
    lop(i,n)
    {
        cin>>a[i]>>b[i]>>c[i];
    }
    ms(mem,-1);
    int ans = dp (0,0,0) ;
    if (ans == 5000 ) ans = -1;
    cout << ans ;
}
/*
 
 
 
https://ideone.com/goT6Vy//http:
codeforces.com/problemset/problem/466/D
 
, vector < pair <int,int > >, greater <pair <int, int> >
http://codeforces.com/contest/295/submission/17929453
freopen("output.txt","w",stdout);
freopen("input.txt","r",stdin);
 
ll pwmod(ll a, ll n, ll mod)
{
    ll ret = 1;
    while (n)
    {
        if (n & 1) ret = ret * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ret;
}
 
for (int i=0; i<n; i++)            Hi    ^_^  hack me if you can :P
 
int mx[]= {0,0,1,-1, 1,1,-1,-1};
int my[]= {1,-1,0,0, 1,-1,1,-1};
bool inside (int i,int j){return (i>=0 && i<n && j>=0 && j<m);}
 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
 
void gcd(LL n,LL m,LL &d,LL &xx,LL &yy)
{
    if(!m)
    {
        d=n;
        xx=1;
        yy=0;
    }
    else
    {
        gcd(m,n%m,d,yy,xx);
        yy-=xx*(n/m);
    }
}
 
int getm(int l,int r)
{
    if(l>r)
        return N;
    if(l==r)
        return 0;
    if(l==r-1)
        return s[l] != s[r];
    if(s[l]==s[r])
        return getm(l+1,r-1);
    else
        return 1+min(getm(l+1,r),getm(l,r-1));
}
 
int BIT[si], a[si], n;
void update(int x, int delta)
{
    for(; x <= N; x += x&-x)
        BIT[x] += delta;
}
int query(int x)
{
    int sum = 0;
    for(; x > 0; x -= x&-x)
        sum += BIT[x];
    return sum;
}
 
#define sc(x)      scanf("%d",&x)
#define scl(x)     scanf("%lld",&x)
#define pr(x)      printf("%d ",x)
#define prl(x)     printf("%lld ",x)
#define prln(x)    printf("%d\n",x)
#define prlln(x)   printf("%lld\n",x)
#define scd(x)     scanf("%lf",&x)
#define prd(x)     printf("%.10f\n",x)
 
 
*/

Measured memory_bytes: 25984