#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  vector<long long int> p(n), q(n);
  for (long long int i = 0; i <= n - 1; i++) cin >> p[i];
  for (long long int i = 0; i <= n - 1; i++) cin >> q[i];
  set<long long int> p1, q1;
  vector<long long int> chan(n + 1, 0);
  long long int cou = 1;
  for (long long int i = 0; i <= n - 1; i++) {
    if (p1.find(q[i]) != p1.end())
      p1.erase(p1.find(q[i]));
    else
      q1.insert(q[i]);
    if (q1.find(p[i]) != q1.end())
      q1.erase(q1.find(p[i]));
    else
      p1.insert(p[i]);
    if (p1.empty() && q1.empty()) {
      chan[i + 1] = cou;
      cou++;
    }
  }
  for (long long int i = 1; i <= n - 1; i++) {
    chan[i] = max(chan[i], chan[i - 1]);
    chan[i] = min(chan[i], k - 1);
  }
  if (chan[n - 1] == k - 1) {
    cout << "YES" << '\n';
    vector<char> ans(n);
    for (long long int i = 0; i <= n - 1; i++) {
      ans[p[i] - 1] = 'a' + chan[i];
    }
    for (long long int i = 0; i <= n - 1; i++) cout << ans[i];
  } else {
    cout << "NO" << '\n';
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  vector<long long int> p(n), q(n);
  for (long long int i = 0; i <= n - 1; i++) cin >> p[i];
  for (long long int i = 0; i <= n - 1; i++) cin >> q[i];
  set<long long int> p1, q1;
  vector<long long int> chan(n + 1, 0);
  long long int cou = 1;
  for (long long int i = 0; i <= n - 1; i++) {
    if (p1.find(q[i]) != p1.end())
      p1.erase(p1.find(q[i]));
    else
      q1.insert(q[i]);
    if (q1.find(p[i]) != q1.end())
      q1.erase(q1.find(p[i]));
    else
      p1.insert(p[i]);
    if (p1.empty() && q1.empty()) {
      chan[i + 1] = cou;
      cou++;
    }
  }
  for (long long int i = 1; i <= n - 1; i++) {
    chan[i] = max(chan[i], chan[i - 1]);
    chan[i] = min(chan[i], k - 1);
  }
  if (chan[n - 1] == k - 1) {
    cout << "YES" << '\n';
    vector<char> ans(n);
    for (long long int i = 0; i <= n - 1; i++) {
      ans[p[i] - 1] = 'a' + chan[i];
    }
    for (long long int i = 0; i <= n - 1; i++) cout << ans[i];
  } else {
    cout << "NO" << '\n';
  }
  return 0;
}

```