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;
const int INF = 0x3f3f3f3f;
const int maxn = 2e6 + 10;
char a[1000];
int main() {
  map<char, int> mmp;
  int k;
  scanf("%d %s", &k, a);
  int len = strlen(a);
  int cur = 0;
  int d[1000];
  if (k == 1) {
    printf("YES\n");
    printf("%s", a);
  } else if (len < k)
    printf("NO\n");
  else {
    for (int i = 0; i < len; i++) {
      if (mmp[a[i]] == 0) {
        mmp[a[i]] = 1;
        d[cur++] = i;
      }
      if (cur == k) break;
    }
    if (cur != k)
      printf("NO\n");
    else {
      printf("YES\n");
      int be = d[0];
      for (int i = 1; i < cur; i++) {
        for (int j = be; j < d[i]; j++) printf("%c", a[j]);
        printf("\n");
        be = d[i];
      }
      for (int i = be; i < len; i++) printf("%c", a[i]);
      printf("\n");
    }
  }
  return 0;
}
