F - Set of Strings

You are given a string q. A sequence of k strings s1, s2, ..., sk is called beautiful, if the concatenation of these strings is string q (formally, s1 + s2 + ... + sk = 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, ..., sk.

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

Examples

Input

1
abca

Output

YES
abca

Input

2
aaacas

Output

YES
aaa
cas

Input

4
abc

Output

NO

Note

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

#include<iostream>

#include<string>

#include<cstring>

#include<map>

#include<queue>

using namespace std;

map<char,int >wakaka;

int a[1010];

int main()

{

    int n;

    cin>>n;

    memset(a,0,sizeof(a));

    string s;

    cin>>s;

    int len=s.size();

    if(len<n)

    {

        cout<<"NO"<<endl;

    }

    else

    {

        a[0]=0;

        int ans=0;

        for(int i=0; i<len; i++)

        {

            if(wakaka[s[i]]==0)

            {

                ans++;

                a[ans]=i;

            }

            wakaka[s[i]]++;

        }

        if(ans>=n)

        {

            cout<<"YES"<<endl;

            for(int i=1; i<=ans; i++)

            {

                if(i==n)

                {

                    for(int j=a[i]; j<len; j++)

                    {

                        cout<<s[j];

                    }

                    break;

                }

                for(int j=a[i]; j<a[i+1]; j++)

                {

                    cout<<s[j];

                }

                cout<<endl;

            }

        }

        else

        {

            cout<<"NO"<<endl;

        }

    }

    return 0;

}