A

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

You are given a string q. A sequence of k strings s₁, s₂, ..., s_k is called beautiful, if the concatenation of these strings is string q(formally, s₁ + s₂ + ... + s_k = q) and the first characters of these strings are distinct.

Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.

Input

The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence.

The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.

Output

If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next k lines print the beautiful sequence of strings s₁, s₂, ..., s_k.

If there are multiple possible answers, print any of them.

Sample Input

Input

1
abca

Output

YES
abca

Input

2
aaacas

Output

YES
aaa
cas

Input

4
abc

Output

NO

Hint

In the second sample there are two possible answers: {"aaaca", "s"} and {"aaa", "cas"}.

题意：

给你k和一个字符串q，求能否将q分为k份且每一份的首字符各不相同。

可用map来筛选不同的字母的个数，可分为k段则至少有k个不同的字母。输出时再用map记录以用过的首字母。

附AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<algorithm>
using namespace std;

int main(){
    string s,a;
    int k,lens,lenm;
    cin>>k>>s;
    int t=1,temp=0,ans=0;
    map<char,int> m;
    lens=s.size();
    for(int i=0;i<lens;i++){
        m[s[i]]=1;
    }
    lenm=m.size();
    a[0]=s[0];
    if(lens>=k&&lenm>=k){
        m.clear();
        cout<<"YES"<<endl;
            for(int i=0;i<lens;i++){
                if(m[s[i]]==0 && temp<k){
                if(temp) cout<<endl;
                temp++;
            }
            m[s[i]]++;
            cout<<s[i];
            }
    }
    else{
        cout<<"NO"<<endl;
    }
    return 0;
}