#include <bits/stdc++.h>
#include <chrono>
using namespace std;
using namespace chrono;
// "AJEET JAIN"----"JAI JINENDRA"
/* "णमो अरिहंताणं",
"णमो सिद्धाणं",
"णमो आयरियाणं",
"णमो उवज्झायाणं",
"णमो लोए सव्वसाहूणं",
"",
"एसो पंच नमोक्कारो, सव्व पावप्पणासणो",
"मंगलाणं च सव्वेसिं, पडमं हवै मंगलं", */
// Aliases to op
using ll = long long;
using ull = unsigned long long;
using ld = double;
using vll = vector<ll>;
// Constants
constexpr ll INF = 4e18;
constexpr ld EPS = 1e-9;
constexpr ll MOD = 1e9 + 7;
// Macros
#define F first
#define S second
#define all(x) begin(x), end(x)
#define allr(x) rbegin(x), rend(x)
#define py cout<<"YES\n";
#define pn cout<<"NO\n";
#define forn(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
// #define insert push_back
#define pb push_back
#define MP make_pair
#define endl '\n'
/*
remove substring or subarray ---> try to think about sliding w
*/
/*
Golden Rule
1) problem is easy
2) proofs is easy
3) implementation is easy
/*
ROUGH --
*/
void AJNJ(){
ll n,k;
cin>>n>>k;
ll arr[n][3];
for(int i = 0 ; i < n ; i++){
for(int j = 0 ; j < 3 ; j++){
cin >> arr[i][j];
}
}
ll sum_a = 0 , sum_b = 0;
for(int i = 0 ; i < n ; i++){
sum_a += arr[i][1];
sum_b += arr[i][2];
}
if(sum_a < k || sum_b < k){
cout << "-1" << endl;
return;
}
vector<ll> v_a , v_b , v_a_b;
for(int i = 0 ; i < n ; i++){
if(arr[i][1] == 1 && arr[i][2] == 0){
v_a.push_back(arr[i][0]);
}
if(arr[i][1] == 0 && arr[i][2] == 1){
v_b.push_back(arr[i][0]);
}
if(arr[i][1] == 1 && arr[i][2] == 1){
v_a_b.push_back(arr[i][0]);
}
}
sort(all(v_a));
sort(all(v_b));
sort(all(v_a_b));
for(int i = 0 ; i < min(v_a.size() , v_b.size()) ; i++){
v_a_b.push_back(v_a[i] + v_b[i]);
}
sort(all(v_a_b));
ll ans = 0;
for(int i = 0 ; i < v_a_b.size() ; i++){
ans += v_a_b[i];
if(i + 1 == k){
break;
}
}
cout << ans << endl;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int T = 1;
// cin>>T;
auto start1 = high_resolution_clock::now();
while(T--){
AJNJ();
}
auto stop1 = high_resolution_clock::now();
auto duration = duration_cast<microseconds>(stop1 - start1);
cerr << "Time: " << duration . count() / 1000 << " ms" << endl;
return 0;
}
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