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 MAXN = (int)1e5 + 123;
double Time() { return (double(clock()) / double(CLOCKS_PER_SEC)); }
int n;
string s, cur;
map<char, int> used;
vector<string> ans;
void f() {
  cout << "YES\n";
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << "\n";
}
void big() {
  cout << "YES\n";
  int i = 0;
  for (; i < ans.size(); i++) {
    if (n > 1)
      n--, cout << ans[i] << "\n";
    else if (n == 1)
      cout << ans[i];
  }
}
int main() {
  cin >> n;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (!used[s[i]]) {
      used[s[i]]++;
      if (cur.size() > 0) ans.push_back(cur);
      cur = "";
    }
    cur += s[i];
  }
  if (cur.size() > 0) ans.push_back(cur);
  if (ans.size() < n)
    cout << "NO";
  else if (ans.size() > n)
    big();
  else if (ans.size() == n)
    f();
  return 0;
}
