#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  vector<int> e(n);
  vector<int> to(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
    e[i] = a[i];
    to[a[i]] = i;
    a[i] = i;
  }
  vector<int> b(n);
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    --b[i];
    b[i] = to[b[i]];
  }
  int min_suf = (int)1e9;
  vector<int> dp(n + 1, 0);
  for (int i = n - 1; i >= 0; --i) {
    if (min_suf < b[i]) {
      ++dp[min_suf + 1];
      --dp[b[i] + 1];
    }
    min_suf = min(min_suf, b[i]);
  }
  for (int i = 1; i < n; ++i) dp[i] += dp[i - 1];
  char last = 'a' - 1;
  string ans = "";
  for (int i = 0; i < n; ++i) {
    if (dp[i] == 0 && last != 'z') ++last;
    ans += last;
  }
  if (last - 'a' + 1 < m) {
    cout << "NO\n";
    return 0;
  }
  string normal_ans = ans;
  for (int i = 0; i < n; ++i) normal_ans[e[i]] = ans[i];
  cout << "YES\n";
  cout << normal_ans << "\n";
}

### 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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  vector<int> e(n);
  vector<int> to(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
    e[i] = a[i];
    to[a[i]] = i;
    a[i] = i;
  }
  vector<int> b(n);
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    --b[i];
    b[i] = to[b[i]];
  }
  int min_suf = (int)1e9;
  vector<int> dp(n + 1, 0);
  for (int i = n - 1; i >= 0; --i) {
    if (min_suf < b[i]) {
      ++dp[min_suf + 1];
      --dp[b[i] + 1];
    }
    min_suf = min(min_suf, b[i]);
  }
  for (int i = 1; i < n; ++i) dp[i] += dp[i - 1];
  char last = 'a' - 1;
  string ans = "";
  for (int i = 0; i < n; ++i) {
    if (dp[i] == 0 && last != 'z') ++last;
    ans += last;
  }
  if (last - 'a' + 1 < m) {
    cout << "NO\n";
    return 0;
  }
  string normal_ans = ans;
  for (int i = 0; i < n; ++i) normal_ans[e[i]] = ans[i];
  cout << "YES\n";
  cout << normal_ans << "\n";
}

```