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 fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int MAXN = 256;
int a[MAXN];
char str[119];
vector<int> v;
int main() {
  int n;
  scanf("%d", &n);
  scanf("%s", str);
  int count = 0;
  int slen = strlen(str);
  memset(a, 0, sizeof a);
  for (int i = 0; i < slen; i++) {
    if (a[str[i]] == 0) {
      count++;
      a[str[i]]++;
      v.push_back(i);
    }
  }
  if (n <= count) {
    printf("YES\n");
    int j = 1;
    count = 1;
    for (int i = 0; i < slen; i++) {
      if (v[j] == i && n > count) {
        printf("\n");
        printf("%c", str[i]);
        count++;
        j++;
      } else {
        printf("%c", str[i]);
      }
    }
    printf("\n");
  } else {
    printf("NO\n");
  }
  return 0;
}
