#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  vector<int> b(n);
  for (int &i : a) cin >> i;
  for (int &i : b) cin >> i;
  map<int, bool> a_map;
  map<int, bool> b_map;
  vector<pair<int, int> > blocks;
  int last = 0;
  for (int i = 0; i < n; ++i) {
    a_map[a[i]] = 1;
    b_map[b[i]] = 1;
    if (b_map[a[i]]) {
      a_map.erase(a[i]);
      b_map.erase(a[i]);
    }
    if (a[i] != b[i] && a_map[b[i]]) {
      a_map.erase(b[i]);
      b_map.erase(b[i]);
    }
    if (a_map.empty()) {
      blocks.push_back({last, i});
      last = i + 1;
    }
  }
  if (blocks.size() < k) return cout << "NO", 0;
  vector<char> ans(n);
  for (int i = 0; i < blocks.size(); ++i) {
    for (int j = blocks[i].first; j <= blocks[i].second; ++j) {
      if (i > 25)
        ans[a[j] - 1] = 'z';
      else
        ans[a[j] - 1] = 'a' + i;
    }
  }
  cout << "YES\n";
  for (char &i : ans) cout << i;
  return 0;
}

### Prompt

Develop a solution in cpp 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() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  vector<int> b(n);
  for (int &i : a) cin >> i;
  for (int &i : b) cin >> i;
  map<int, bool> a_map;
  map<int, bool> b_map;
  vector<pair<int, int> > blocks;
  int last = 0;
  for (int i = 0; i < n; ++i) {
    a_map[a[i]] = 1;
    b_map[b[i]] = 1;
    if (b_map[a[i]]) {
      a_map.erase(a[i]);
      b_map.erase(a[i]);
    }
    if (a[i] != b[i] && a_map[b[i]]) {
      a_map.erase(b[i]);
      b_map.erase(b[i]);
    }
    if (a_map.empty()) {
      blocks.push_back({last, i});
      last = i + 1;
    }
  }
  if (blocks.size() < k) return cout << "NO", 0;
  vector<char> ans(n);
  for (int i = 0; i < blocks.size(); ++i) {
    for (int j = blocks[i].first; j <= blocks[i].second; ++j) {
      if (i > 25)
        ans[a[j] - 1] = 'z';
      else
        ans[a[j] - 1] = 'a' + i;
    }
  }
  cout << "YES\n";
  for (char &i : ans) cout << i;
  return 0;
}

```