#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 15;
int n, k, a[N], b[N], res[N], qid;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0, ggd = n; i < ggd; i++) cin >> a[i];
  for (int i = 0, ggd = n; i < ggd; i++) cin >> b[i];
  set<int> l, r, cur;
  for (int i = 0, ggd = n; i < ggd; i++) {
    cur.insert(a[i]);
    cur.insert(b[i]);
    if (a[i] != b[i]) {
      if (l.count(a[i]))
        l.erase(a[i]);
      else
        r.insert(a[i]);
      if (r.count(b[i]))
        r.erase(b[i]);
      else
        l.insert(b[i]);
    }
    if (!(((int)(l).size()))) {
      assert(!(((int)(r).size())));
      for (auto it : cur) res[it] = qid;
      cur.clear();
      qid++;
    }
  }
  if (((int)(cur).size())) {
    for (auto it : cur) res[it] = qid;
    qid++;
  }
  if (qid >= k) {
    cout << "YES\n";
    for (int i = 1, ggd = n + 1; i < ggd; i++)
      cout << (char)('a' + min(25, res[i]));
    cout << "\n";
  } else
    cout << "NO\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Authors have come up with the string s consisting of n lowercase Latin letters.

You are given two permutations of its indices (not necessary equal) p and q (both of length n). Recall that the permutation is the array of length n which contains each integer from 1 to n exactly once.

For all i from 1 to n-1 the following properties hold: s[p_i] ≤ s[p_{i + 1}] and s[q_i] ≤ s[q_{i + 1}]. It means that if you will write down all characters of s in order of permutation indices, the resulting string will be sorted in the non-decreasing order.

Your task is to restore any such string s of length n consisting of at least k distinct lowercase Latin letters which suits the given permutations.

If there are multiple answers, you can print any of them.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 26) — the length of the string and the number of distinct characters required.

The second line of the input contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct integers from 1 to n) — the permutation p.

The third line of the input contains n integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n, all q_i are distinct integers from 1 to n) — the permutation q.

Output

If it is impossible to find the suitable string, print "NO" on the first line.

Otherwise print "YES" on the first line and string s on the second line. It should consist of n lowercase Latin letters, contain at least k distinct characters and suit the given permutations.

If there are multiple answers, you can print any of them.

Example

Input


3 2
1 2 3
1 3 2


Output


YES
abb

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 15;
int n, k, a[N], b[N], res[N], qid;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0, ggd = n; i < ggd; i++) cin >> a[i];
  for (int i = 0, ggd = n; i < ggd; i++) cin >> b[i];
  set<int> l, r, cur;
  for (int i = 0, ggd = n; i < ggd; i++) {
    cur.insert(a[i]);
    cur.insert(b[i]);
    if (a[i] != b[i]) {
      if (l.count(a[i]))
        l.erase(a[i]);
      else
        r.insert(a[i]);
      if (r.count(b[i]))
        r.erase(b[i]);
      else
        l.insert(b[i]);
    }
    if (!(((int)(l).size()))) {
      assert(!(((int)(r).size())));
      for (auto it : cur) res[it] = qid;
      cur.clear();
      qid++;
    }
  }
  if (((int)(cur).size())) {
    for (auto it : cur) res[it] = qid;
    qid++;
  }
  if (qid >= k) {
    cout << "YES\n";
    for (int i = 1, ggd = n + 1; i < ggd; i++)
      cout << (char)('a' + min(25, res[i]));
    cout << "\n";
  } else
    cout << "NO\n";
}

```