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, all = 0, l;
char s[104];
int vis[104];
struct data {
  bool t;
  int a, x;
} used[104];
int p[34];
inline bool dex(data a, data b) { return a.x < b.x; }
int main() {
  memset(used, 0, sizeof(used));
  scanf("%d", &k);
  scanf("%s", s + 1);
  l = strlen(s + 1);
  for (int i = 1; i <= 26; i++) used[i].a = i;
  for (int i = 1; i <= l; i++)
    if (!used[s[i] - 'a' + 1].t) {
      used[s[i] - 'a' + 1].t = 1;
      used[s[i] - 'a' + 1].x = i;
    }
  sort(used + 1, used + 26 + 1, dex);
  for (int i = 1; i <= 26; i++) {
    if (used[i].t == 1) {
      vis[++all] = used[i].x;
    }
  }
  if (all < k)
    return puts("NO"), 0;
  else
    puts("YES");
  vis[all + 1] = l;
  for (int i = 1; i < k; i++) {
    for (int j = vis[i]; j < vis[i + 1]; j++) cout << s[j];
    cout << endl;
  }
  for (int i = vis[k]; i <= l; i++) cout << s[i];
  return 0;
}
