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 main() {
  long long int n, m, i, j, k, cnt = 0, sum = 0, mn = INT_MAX, mx = INT_MIN,
                               diff, ans, num, x, y, z, size, ee, ind;
  int flag = 0, t = 1;
  string s, str, s1, s2, s3;
  char ch;
  while (t--) {
    cin >> k >> s;
    if (k == 1) {
      cout << "YES" << endl << s;
      return 0;
    }
    n = s.size();
    vector<int> pos;
    unordered_set<char> st;
    st.insert(s[0]);
    pos.push_back(0);
    for (i = 0; i < n; i++) {
      if (st.count(s[i]) == 0) {
        pos.push_back(i);
        st.insert(s[i]);
        if (pos.size() == k) break;
      }
    }
    if (pos.size() < k) {
      cout << "NO";
      return 0;
    }
    ind = 0;
    cout << "YES" << endl;
    for (i = 0; i < k; i++) {
      if (i == k - 1) {
        for (j = pos[i]; j < n; j++) cout << s[j];
      } else {
        while (ind < pos[i + 1]) {
          cout << s[ind];
          ind++;
        }
        cout << endl;
      }
    }
  }
}
