#include <bits/stdc++.h>
using namespace std;
long long const N = 1e6 + 5, mod = 1e9 + 7;
long long n, m, k, i, j, kaskad[N], dp[N], mas[N], minn, maxx;
pair<long long, long long> ans[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin.tie(0);
  cin >> n >> k;
  for (i = 1; i <= n; ++i) {
    cin >> kaskad[i];
  }
  string s;
  for (i = 1; i <= n; ++i) {
    ans[i] = make_pair(kaskad[i], i);
    s += '0';
  }
  sort(ans + 1, ans + n + 1);
  for (i = 1; i <= n; ++i) cin >> mas[i];
  set<pair<long long, long long> > my_set;
  for (i = 0; i < n; ++i) my_set.insert(make_pair(i + 1, i + 1));
  for (i = 1; i <= n; ++i) {
    minn = min(i, ans[mas[i]].second);
    maxx = max(i, ans[mas[i]].second);
    auto it = my_set.lower_bound(make_pair(minn, mod));
    it--;
    pair<long long, long long> p = *it;
    it++;
    while (it != my_set.end() && it->first <= maxx) {
      p.second = it->second;
      auto it1 = it;
      it++;
      my_set.erase(it1);
    }
    it--;
    my_set.erase(it);
    my_set.insert(p);
  }
  if (my_set.size() < k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  long long SIZE = my_set.size();
  auto it = my_set.begin();
  char chr = 'a';
  for (i = 1; i <= n; ++i) {
    if (k <= 1) {
      s[kaskad[i] - 1] = chr;
      continue;
    }
    for (i = it->first; i <= it->second; ++i) {
      s[kaskad[i] - 1] = chr;
    }
    i--;
    chr++;
    it++;
    k--;
  }
  cout << s << endl;
  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;
long long const N = 1e6 + 5, mod = 1e9 + 7;
long long n, m, k, i, j, kaskad[N], dp[N], mas[N], minn, maxx;
pair<long long, long long> ans[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin.tie(0);
  cin >> n >> k;
  for (i = 1; i <= n; ++i) {
    cin >> kaskad[i];
  }
  string s;
  for (i = 1; i <= n; ++i) {
    ans[i] = make_pair(kaskad[i], i);
    s += '0';
  }
  sort(ans + 1, ans + n + 1);
  for (i = 1; i <= n; ++i) cin >> mas[i];
  set<pair<long long, long long> > my_set;
  for (i = 0; i < n; ++i) my_set.insert(make_pair(i + 1, i + 1));
  for (i = 1; i <= n; ++i) {
    minn = min(i, ans[mas[i]].second);
    maxx = max(i, ans[mas[i]].second);
    auto it = my_set.lower_bound(make_pair(minn, mod));
    it--;
    pair<long long, long long> p = *it;
    it++;
    while (it != my_set.end() && it->first <= maxx) {
      p.second = it->second;
      auto it1 = it;
      it++;
      my_set.erase(it1);
    }
    it--;
    my_set.erase(it);
    my_set.insert(p);
  }
  if (my_set.size() < k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  long long SIZE = my_set.size();
  auto it = my_set.begin();
  char chr = 'a';
  for (i = 1; i <= n; ++i) {
    if (k <= 1) {
      s[kaskad[i] - 1] = chr;
      continue;
    }
    for (i = it->first; i <= it->second; ++i) {
      s[kaskad[i] - 1] = chr;
    }
    i--;
    chr++;
    it++;
    k--;
  }
  cout << s << endl;
  return 0;
}

```