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 <bits/stdc++.h>
using namespace std;
void read(long long &n) { scanf("%lld", &n); }
void read(double &n) { scanf("%lf", &n); }
void read(int &n) { scanf("%d", &n); }
void read(char *s) { scanf("%s", s); }
template <class T>
inline T bigMod(T p, T e, T M) {
  long long ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
    e >>= 1;
  }
  return (T)ret;
}
template <class T>
inline T power(T p, T e) {
  long long ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
    e >>= 1;
  }
  return (T)ret;
}
int main() {
  int n, vis[10000];
  memset(vis, 0, sizeof(vis));
  ;
  string s;
  cin >> n >> s;
  set<char> se(s.begin(), s.end());
  if (se.size() >= n) {
    cout << "YES" << endl;
    int f = 0;
    for (auto p : s) {
      if (vis[p] == 0 && n) {
        if (f) cout << endl;
        f = 1;
        cout << p;
        n -= 1;
        vis[p] = 1;
      } else
        cout << p;
    }
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
