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;
int n, len;
string cad;
int car[1000];
bool flag[1000];
int sol(int pos) {
  for (int i = pos; i <= len; i++) {
    if (!flag[cad[i]]) {
      flag[cad[i]] = 1;
      return i;
    }
  }
  return -1;
}
int main() {
  scanf("%d", &n);
  cin >> cad;
  len = cad.length();
  int cant = 0;
  for (int i = 0; i < len; i++) {
    if (car[cad[i]] == 0) {
      cant++;
    }
    car[cad[i]]++;
  }
  if (cant >= n) {
    printf("YES\n");
    int x = 0;
    int ant = 0;
    flag[cad[0]] = 1;
    for (int i = 0; i < n - 1; i++) {
      ant = x;
      x = sol(x);
      cout << cad.substr(ant, x - ant) << endl;
    }
    cout << cad.substr(x, len - x) << endl;
  } else {
    printf("NO\n");
  }
  return 0;
}
