/*

https://yukicoder.me/problems/no/1029
個数無制限ナップザック？
部分文字列としてどこまで完成されているかを
0 ～ 3Kで表す?

各Kに関して、どの数字の時に何を足すといくつ増えるかをO(1)でだせれば
あとはO(KN)の平易なdpでいける

適当に2分探索すれば行けるのでは？？
*/

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n,m) for(ll (i)=(n);(i)<(m);(i)++)
#define rrep(i,n,m) for(ll (i)=(n);(i)>(m);(i)--)
using ll = long long;
const ll mod = 998244353;

ll want(ll now,ll K,string& S){

    rep(i,0,S.size()){
        if ((now < K) && (S[i] == 'J')){
            now++;
        }else if( (K <= now)&&(now < 2*K)&&(S[i]=='O' )){
            now++;
        }else if( (2*K <= now) && (now < 3*K) && (S[i]=='I')){
            now++;
        }
    }

    return now;
}

int main(){

    ll N,K;
    cin >> N >> K;
    int flag = 0;

    vector<string> S(N,"");
    vector<ll> C(N,0);

    rep(i,0,N){
        cin >> S[i] >> C[i];
    }

    vector<ll> JN(N,0);
    vector<ll> ON(N,0);
    vector<ll> IN(N,0);

    rep(i,0,N){
        rep(j,0,S[i].size()){
            if (S[i][j] == 'J') JN[i]++;
            if (S[i][j] == 'O') ON[i]++;
            if (S[i][j] == 'I') IN[i]++;
        }
    }

    vector<ll> ans(3*K+1,LLONG_MAX/5);
    ans[0] = 0;

    ll nex;
    rep(i,0,3*K){
        rep(j,0,N){
            
            if (i+JN[j] < K-1){
                nex = i+JN[j];
            }else if ( (K<=i) && (i+ON[j] < 2*K-1)){
                nex = i+ON[j];
            }else if ( i >= 2*K ){
                nex = min(3*K,i+IN[j]);
            }else{
                nex = want(i,K,S[j]);
            }
            ans[nex] = min(ans[nex] , ans[i] + C[j]);
            if (nex == 3*K) flag = 1;

        }
    }

    if ((flag == 1) && (ans[3*K] != LLONG_MAX/5)){
        cout << ans[3*K] << endl;
    }else{
        cout << -1 << endl;
    }

}