#include <bits/stdc++.h>
using namespace std;
int s[200005];
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> p1, p2;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    p1.push_back(a);
  }
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    p2.push_back(a);
  }
  set<int> pos1, pos2;
  vector<int> save;
  int cur = 0;
  for (int i = 0; i < n; i++) {
    if (pos2.find(p1[i]) != pos2.end())
      pos2.erase(pos2.find(p1[i]));
    else
      pos1.insert(p1[i]);
    if (pos1.find(p2[i]) != pos1.end())
      pos1.erase(pos1.find(p2[i]));
    else
      pos2.insert(p2[i]);
    save.push_back(p1[i]);
    save.push_back(p2[i]);
    if (pos1.empty() && pos2.empty()) {
      while (save.size()) {
        s[save.back() - 1] = cur;
        save.pop_back();
      }
      cur++;
    }
  }
  if (cur < k)
    cout << "NO";
  else {
    cout << "YES" << '\n';
    for (int i = 0; i < n; i++) {
      if (s[i] <= 25)
        cout << (char)(s[i] + 'a');
      else
        cout << 'z';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int s[200005];
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> p1, p2;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    p1.push_back(a);
  }
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    p2.push_back(a);
  }
  set<int> pos1, pos2;
  vector<int> save;
  int cur = 0;
  for (int i = 0; i < n; i++) {
    if (pos2.find(p1[i]) != pos2.end())
      pos2.erase(pos2.find(p1[i]));
    else
      pos1.insert(p1[i]);
    if (pos1.find(p2[i]) != pos1.end())
      pos1.erase(pos1.find(p2[i]));
    else
      pos2.insert(p2[i]);
    save.push_back(p1[i]);
    save.push_back(p2[i]);
    if (pos1.empty() && pos2.empty()) {
      while (save.size()) {
        s[save.back() - 1] = cur;
        save.pop_back();
      }
      cur++;
    }
  }
  if (cur < k)
    cout << "NO";
  else {
    cout << "YES" << '\n';
    for (int i = 0; i < n; i++) {
      if (s[i] <= 25)
        cout << (char)(s[i] + 'a');
      else
        cout << 'z';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}

```