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
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>
#include <ctime>
using namespace std;
 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
 

//========================TypeDefs===================================
typedef int elem_t;
typedef vector<long long> vll;
typedef vector<vector<long long>> vvl;
typedef vector<int> vi; 
typedef vector<vi> vvi; 
typedef pair<int,int> ii; 

 
//=========================MACROS====================================

#define shen01orz() main()
#define max(a,b) (a<b?b:a)
#define min(a,b) (a>b?b:a)
#define rep(i,n) for (i = 0; i < n; ++i) 
#define REP(i,k,n) for (i = k; i <= n; ++i) 
#define REPR(i,k,n) for (i = k; i >= n; --i)
#define clr(a,x) memset(a,x,sizeof(a)) //set elements of array to some value
#define all(c) (c).begin(),(c).end() //handy for function like "sort()"
#define ll long long //data types used often, but you don't want to type them time by time
#define IOS ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0);//to synchronize the input of cin and scanf
#define INF (long long)1e18
#define intINF (int)1001001001
#define PI 3.1415926535897932384626
//for map, pair
#define mp make_pair
#define fi first
#define se second
#define pb push_back
// directions
const int fx[4][2] = {{0,1}, {-1,0}, {0,-1}, {1,0}};
const int fxx[9][2] = {{0,0},{0,1}, {0,-1}, {1,0}, {-1,0}, {1,1}, {1,-1}, {-1,1}, {-1,-1}};
const ll mod = 1000000007;
#define deci(n) fixed << setprecision(n)
#define high(n) __builtin_popcount(n)
#define highll(n) __builtin_popcountll(n)
#define parity(n) __builtin_parity(n)
#define clz(n) __builtin_clz(n)
#define clzll(n) __builtin_clzll(n)
#define ctz(n) __builtin_ctz(n)

//oset
#define oset tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define osetlli tree<lli, null_type, less<lli>, rb_tree_tag, tree_order_statistics_node_update>
 
#define ofk order_of_key
#define fbo find_by_order

 
//==============================FUNCTIONS===========================================
inline ll power(ll x, ll y, ll p = mod)
{
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
 
inline ll modadd(ll a, ll b, ll m = mod)
{
    a += b;
    if (a >= m)
        a -= m;
    return a;
}
 
inline ll modmul(ll a, ll b, ll m = mod)
{
    return ((a % m) * (b % m)) % m;
}
 
inline ll modi(ll a, ll m = mod) { return power(a, m - 2, m); }
 
long long int inverse(long long int i,ll m=mod){
    if(i==1) return 1;
    return (m - ((m/i)*inverse(m%i))%m+m)%m;
}

ll rand_int(ll l, ll r) 
{ //[l, r]
    #ifdef LOCAL
    static mt19937_64 gen;
    #else
    static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
    #endif
    return uniform_int_distribution<ll>(l, r)(gen);
}

void solve()
{
    ll n,l,k;
    cin>>n>>l>>k;
    string s;
    cin>>s;
    vi curr;
    for(ll i=0;i<n*l;i++)
    {
        int x;
        x = s[i];
        curr.pb(x);
    }
    sort(all(curr));
    ll ed = n*l - 1;
    for(ll i = 0;i<k-1;i++)
    {
        char x;
        x = curr[i];
        cout<<x;
        for(ll j = ed;j>ed-l + 1;j--)
        {
            x = curr[j];
            cout<<x;
        }
        ed-=(l-1);
        cout<<"\n";
    }
    for(ll i = 0;i<=n-k;i++)
    {
        char x;
        for(ll j = 0;j<l;j++)
        {
            x = curr[k - 1 + i*l + j];
            cout<<x;
        }
        cout<<"\n";
    }
    
}

elem_t shen01orz()
{
    long long t=1;
    while(t--)
    {
        solve();
    }
}

