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;
queue<string> q;
string str, ans[30], tmp;
bool vis[300];
int main() {
  int index, m;
  int i, j, len, k, cnt;
  while (scanf("%d", &k) != EOF) {
    m = k;
    k--;
    cin >> str;
    len = str.length();
    memset(vis, false, sizeof vis);
    index = i = 0;
    cnt = 0;
    while (1) {
      if (k == 0 || i >= len) break;
      tmp = "";
      vis[str[index]] = true;
      for (; i < len; i++) {
        if (vis[str[i]])
          tmp += str[i];
        else
          break;
      }
      ans[cnt++] = tmp;
      index = i;
      k--;
    }
    if (i < len) {
      tmp = "";
      int t;
      for (; i < len; i++) ans[cnt] += str[i];
      cnt++;
    }
    if (cnt == m) {
      printf("YES\n");
      for (i = 0; i < cnt; i++) cout << ans[i] << endl;
    } else
      printf("NO\n");
  }
  return 0;
}
