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
        #include<bits/stdc++.h>
using namespace std;
typedef long long int lli;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<lli,lli> pll;
typedef std::vector<lli> vll;
typedef std::vector<pll> vpll;
typedef std::vector<bool> vbb;
typedef map<lli,lli>mll;
typedef set<lli> sll;
typedef multiset<lli> msll;
lli MOD=1e9 +7;
lli INF=1e18;
        #define forr(i,p,n) for(i=p; i<n; i++)
        #define forl(i,n,p) for(i=n; i>=p; i--)
        #define be(a) a.begin(),a.end()
        #define mp make_pair
        #define um unordered_map
        #define eb emplace_back 
        #define pb push_back 
        #define F first
        #define S second
        #define all(a)     (a).begin(),(a).end()
        #define PI 3.14159265358979323846
        #define lb lower_bound
        #define ub upper_bound
        #define endl "\n"
    #define deb(vari) cerr<<#vari<<" = "<<vari<<endl;
lli power(lli base, lli exp) {lli res=1;while(exp>0) {if(exp%2==1) res=(res*base);base=(base*base);exp/=2;}return res;}
lli bitc(lli n,lli x){return ((n>>x)&1);}
lli __gcd(lli a,lli b){return b==0?a:__gcd(b,a%b);}
lli fsub(lli a,lli b,lli p=MOD){return ((a%p)-(b%p)+p)%p;}
lli fmult(lli a,lli b, lli p=MOD) {return ((a%p)*(b%p))%p;}
lli fadd(lli a, lli b, lli p=MOD) {return (a%p + b%p)%p;}
        // lli add(lli a,lli b,lli p=MOD){//if(a+b>=p){//return (a+b)-p;//  }//return a+b;// }
lli fpow(lli n, lli k, lli p = MOD) {lli r = 1; while(k > 0) {if (k & 1) r = r * n%p; n = n * n%p; k = k >> 1;} return r;}
lli inv(lli a, lli p = MOD) {return fpow(a, p - 2, p);}
lli fdiv(lli a,lli b,lli p=MOD){lli yinv=inv(b);lli ans=(a*yinv)%p;return ans;}

        template<typename T>istream &operator>>(istream &in, vector<T> &a) {for (auto &item : a) {in >> item;}return in;}template<typename T,typename U>ostream &operator<<(ostream &out, pair<T,U> &a) {cout<<a.F<<" "<<a.S;return out;}
        template<typename T,typename U>
istream &operator>>(istream &out, pair<T,U> &a) {cin>>a.F>>a.S;return out;}
        template<typename T,typename U>
ostream &operator<<(ostream &out, map<T,U> &a) {for (auto &item : a) {out << item << endl;}return out;}
        template<typename T>
ostream &operator<<(ostream &out, vector<T> &a) {for (auto &item : a) {out << item << " ";}return out;}
        template<typename T>
ostream &operator<<(ostream &out, vector<vector<T>> &a) {for (auto &item : a) {out << item << "\n";}return out;}
        #define print(a,c) for(auto p:a) cout<<p<<c; cout<<endl;
        #define read(a) for(auto &p:a)cin>>p>>endl;
void printmat(auto v,lli n,lli m){lli i,j;forr(i,0,n){forr(j,0,m){cout<<v[i][j]<<" ";}cout<<endl;}}
void printarr(auto v,lli n){lli i;forr(i,0,n){cout<<v[i]<<" ";}cout<<endl;}
std::vector<bool> is_prime;
std::vector<lli> primes;
void sieve(lli n){is_prime.resize(n+2,true);primes.clear();lli p;for(p=2;p*p<=n;p++){if(is_prime[p]){lli i;for(i=p*p;i<=n;i+=p){is_prime[i]=false;}}}is_prime[0]=is_prime[1]=false;lli i;for(i=2;i<=n;i++){if(is_prime[i]){primes.eb(i);}}}
mll prime_factors(lli n){  mll s;lli i;lli tc=0;while(n%2==0){tc++;n/=2;}if(tc>0){s[2]=tc;}for(i=3;i<=sqrt(n);i+=2){tc=0;while(n%i==0){tc++;n/=i;}if(tc>0){s[i]=tc;}}if(n>2){s[n]+=1;}return s;}
std::vector<lli> fact_vec;
void fact_fun(lli n){fact_vec.resize(n+10);lli i;fact_vec[0]=1;for(i=1;i<=n+2;i++){fact_vec[i]=(fact_vec[i-1]*i)%MOD;}}
vll p2;
void power_2(lli n,lli m=MOD){ lli i;MOD=m;p2.eb(1);forr(i,0,n){p2.eb(fmult(p2.back(),2));}}
lli ncr(lli n,lli r){if(r>n)return 0;return fdiv(fact_vec[n],fmult(fact_vec[r],fact_vec[n-r]));}


// class Sol{
// public:             
//     Sol(){      
//     }          
// };


int main(){     
    lli i,j;    
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    lli ti;
    // cin>>ti; 
    ti=1;            
    while (ti--){
        lli n,l,k;
        cin>>n>>l>>k;
        string s;
        cin>>s;        
        vector<std::vector<char> >ans(n,std::vector<char> (l,'#'));
        vll cnt(26);
        for(auto p:s){
            cnt[p-'a']++;
        }
        i=0;
        k--;
        j=0;
        lli cur=0;
        lli fs=0;
        while(1){
            if(i>=n){
                break;
            }
            if(j>=l){
                break;
            }
            do{
                while(cnt[cur]==0){
                    fs=i;
                    cur++;
                }   
                ans[i][j]=cur+'a';
                cnt[cur]--;
                i++;
            }while(i<=k);
            if(fs==k){
                break;
            }
            i=fs;
            j++;
            break;
        }
        i=k;
        j++;
        while(j<l){
            while(cnt[cur]==0){
                cur++;
            }
            ans[i][j++]=cur+'a';
            cnt[cur]--;
        }

        forr(i,0,n){
            forr(j,0,l){
                if(ans[i][j]=='#'){
                    while(cnt[cur]==0){
                        cur++;
                    }
                    ans[i][j]=cur+'a';
                    cnt[cur]--;
                }
            }
        }
        for(auto p:ans){
            for(auto q:p){
                cout<<q;
            }
            cout<<endl;
        }


    }
}   