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 k, i, j, l, x = 0;
char a[1000];
bool b[1000];
int main() {
  cin >> k;
  scanf("%s", a);
  l = strlen(a);
  memset(b, true, sizeof(b));
  for (i = 0; i < l; i++)
    if (b[(int)(a[i]) - 97]) {
      ++x;
      b[(int)(a[i]) - 97] = false;
    }
  if (x >= k) {
    x = 0;
    printf("YES\n");
    for (i = 0; i < k - 1; i++) {
      b[(int)(a[x]) - 97] = true;
      while (b[(int)(a[x]) - 97]) {
        printf("%c", a[x]);
        ++x;
      }
      printf("\n");
    }
    for (i = x; i < l; i++) printf("%c", a[i]);
  } else
    printf("NO");
  return 0;
}
