#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v1(n);
  vector<long long int> v2(n);
  for (long long int i = 0; i < n; i++) cin >> v1[i];
  for (long long int i = 0; i < n; i++) cin >> v2[i];
  set<long long int> s1, s2;
  vector<long long int> v;
  v.push_back(-1);
  for (long long int i = 0; i < n; i++) {
    if (s2.find(v1[i]) != s2.end())
      s2.erase(v1[i]);
    else
      s1.insert(v1[i]);
    if (s1.find(v2[i]) != s1.end())
      s1.erase(v2[i]);
    else
      s2.insert(v2[i]);
    if (s1.size() == 0 && s2.size() == 0) {
      v.push_back(i);
    }
  }
  if (v.size() < k + 1) {
    cout << "NO";
    return 0;
  }
  char u = 'a';
  vector<char> c(n + 1);
  for (long long int i = 1; i < v.size(); i++) {
    if (i > 26) {
      long long int j = v[i - 1] + 1;
      while (j <= v[i]) {
        c[v1[j]] = 'z';
        j++;
      }
    } else {
      long long int j = v[i - 1] + 1;
      while (j <= v[i]) {
        c[v1[j]] = u;
        j++;
      }
      u = (char)(u + 1);
    }
  }
  cout << "YES" << endl;
  for (long long int i = 1; i < n + 1; i++) cout << c[i];
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v1(n);
  vector<long long int> v2(n);
  for (long long int i = 0; i < n; i++) cin >> v1[i];
  for (long long int i = 0; i < n; i++) cin >> v2[i];
  set<long long int> s1, s2;
  vector<long long int> v;
  v.push_back(-1);
  for (long long int i = 0; i < n; i++) {
    if (s2.find(v1[i]) != s2.end())
      s2.erase(v1[i]);
    else
      s1.insert(v1[i]);
    if (s1.find(v2[i]) != s1.end())
      s1.erase(v2[i]);
    else
      s2.insert(v2[i]);
    if (s1.size() == 0 && s2.size() == 0) {
      v.push_back(i);
    }
  }
  if (v.size() < k + 1) {
    cout << "NO";
    return 0;
  }
  char u = 'a';
  vector<char> c(n + 1);
  for (long long int i = 1; i < v.size(); i++) {
    if (i > 26) {
      long long int j = v[i - 1] + 1;
      while (j <= v[i]) {
        c[v1[j]] = 'z';
        j++;
      }
    } else {
      long long int j = v[i - 1] + 1;
      while (j <= v[i]) {
        c[v1[j]] = u;
        j++;
      }
      u = (char)(u + 1);
    }
  }
  cout << "YES" << endl;
  for (long long int i = 1; i < n + 1; i++) cout << c[i];
  return 0;
}

```