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;
long long int ipow(long long int base, long long int exp) {
  long long int result = 1;
  while (exp) {
    if (exp & 1) {
      result *= base;
      result %= 1000000007;
    }
    exp >>= 1;
    base *= base;
    base %= 1000000007;
  }
  return result;
}
int main() {
  int t;
  scanf("%d", &t);
  char str[100000];
  scanf("%s", str);
  int len = strlen(str);
  bool done[256];
  memset(done, false, sizeof(done));
  vector<int> z;
  if (t > len)
    printf("%s\n", "NO");
  else {
    for (int i = (int)0; i <= (int)len - 1; i++) {
      if (!done[str[i]]) {
        z.push_back(i);
        done[str[i]] = true;
      }
    }
    if (z.size() < t) {
      printf("%s\n", "NO");
    } else {
      cout << "YES" << endl;
      for (int i = (int)0; i <= (int)t - 2; i++) {
        for (int j = z[i]; j < z[i + 1]; j++) cout << str[j];
        cout << endl;
      }
      for (int j = z[t - 1]; j < len; j++) cout << str[j];
      cout << endl;
    }
  }
  return 0;
}
