Lucy likes letters. She studied the definition of the lexicographical order at school and plays with it.

At first, she tried to construct the lexicographically smallest word out of given letters. It was so easy! Then she tried to build multiple words and minimize one of them. This was much harder!

Formally, Lucy wants to make n words of length l each out of the given n ⋅ l letters, so that the k-th of them in the lexicographic order is lexicographically as small as possible.

Input

The first line contains three integers n, l, and k (1≤ k ≤ n ≤ 1 000; 1 ≤ l ≤ 1 000) — the total number of words, the length of each word, and the index of the word Lucy wants to minimize.

The next line contains a string of n ⋅ l lowercase letters of the English alphabet.

Output

Output n words of l letters each, one per line, using the letters from the input. Words must be sorted in the lexicographic order, and the k-th of them must be lexicographically as small as possible. If there are multiple answers with the smallest k-th word, output any of them.

Examples

Input


3 2 2
abcdef


Output


af
bc
ed

Input


2 3 1
abcabc


Output


aab
bcc
//! Bismillahi-Rahamanirahim.
/** ========================================**
 ** @Author: Md. Abu Farhad ( RUET, CSE'15)
 ** @Category:
/** ========================================**/

#include<bits/stdc++.h>
#include<stdio.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;


#define ll                  long long
#define scl(n)              scanf("%lld",&n)
#define scll(n, m)          scanf("%lld%lld",&n, &m)
#define scc(c)	            scanf("%c",&c)
#define fr(i,n)             for (ll i=0;i<n;i++)
#define fr1(i,n)            for(ll i=1;i<=n;i++)
#define pfl(x)              printf("%lld\n",x)
#define pb                  push_back
#define debug               cout<<"I am here"<<endl;
#define pno                 cout<<"NO"<<endl
#define pys                 cout<<"YES"<<endl
#define tcas(i,t)           for(ll i=1;i<=t;i++)
#define all(x) 	            (x).begin(), (x).end()
#define allrev(x)           (x).rbegin(),(x).rend()
#define pr                  pair<ll, ll>
#define ff                  first
#define ss                  second
#define pn                  printf("\n")

#define l(s)                s.size()
#define asort(a)            sort(a,a+n)
#define dsort(a)            sort(a,a+n,greater<int>())
#define uniquee(x)          x.erase(unique(x.begin(), x.end()),x.end())
#define el                  cout<<endl
#define md                  1000000007
#define inf                 1e18
#define ps                  cout<<" ";
#define Pai                 acos(-1.0)
#define mem(a,i)            memset(a, i, sizeof(a))
#define pcas(i)             cout<<"Case "<<i<<": "<<endl;
#define fast 	ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)

#define N 200006


int main()
{
    ll t;
    t=1;
    tcas(cs, t)
    {
        ll m,n,b,c,d,i,j,k,x,y,z,l,r, p, q;
        string s,s1, s2, s3, s4;

        ll cnt=0,cn=0,sum=0 ;
        cin>>n>>l>>k>>s;

        sort(all(s));

        ll vis[s.size()]={0};

        string ans[s.size()];
        for(i=k-1; i<s.size() ; i++)
        {
             if( s1.size( ) <l )s1+=s[i]  , vis[i]=1;
        }

        //cout<<s1<<endl;

        s2=s1;
        s1.clear();

        for(i=0 , j=0 ; i<s.size()  ; i++ )
        {
               //if(i==k-1)j++;
               if(!vis[i])
               {
                   if(s1.size()<l)  s1+=s[i];
                   else
                   {
                       ans[ j ]=s1 ,
                       s1.clear() , j++;
                       i--;
                   }
               }
        }

        //pfl(j);

        ans[j]=s1;

        //cout<<ans[0]<<" "<<ans[1]<<" "<<ans[2]<<endl;

        fr(i, k-1)cout<<ans[i]<<endl;
        cout<<s2<<endl;
        for(i=k-1;i<n; i++)cout<<ans[i]<<endl;

    }

return 0;
}

/// **************************Before submit****************************

///    ****Please check all base case output  and printing " YES or NO " ***
///    *check for integer overflow,array bounds
///    *check for n=1
