#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n), q(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> p[i];
    --p[i];
  }
  for (int i = 0; i < (int)(n); ++i) {
    cin >> q[i];
    --q[i];
  }
  vector<int> b;
  int s = 0;
  while (s < n) {
    int i = s;
    set<int> P, Q;
    while (i < n) {
      if (p[i] != q[i]) {
        P.insert(p[i]);
        Q.insert(q[i]);
        if (Q.count(p[i])) P.erase(p[i]);
        if (P.count(q[i])) P.erase(q[i]);
      }
      ++i;
      if (P.empty()) break;
    }
    b.push_back(i);
    s = i;
  }
  int B = b.size();
  if (B >= k) {
    string ans(n, '?');
    int pre = 0;
    for (int i = 0; i < (int)(B); ++i) {
      char c = 'a' + min(i, k - 1);
      for (int j = pre; j < b[i]; ++j) ans[p[j]] = c;
      pre = b[i];
    }
    cout << "YES\n";
    cout << ans << "\n";
  } else
    cout << "NO\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ll = long long;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n), q(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> p[i];
    --p[i];
  }
  for (int i = 0; i < (int)(n); ++i) {
    cin >> q[i];
    --q[i];
  }
  vector<int> b;
  int s = 0;
  while (s < n) {
    int i = s;
    set<int> P, Q;
    while (i < n) {
      if (p[i] != q[i]) {
        P.insert(p[i]);
        Q.insert(q[i]);
        if (Q.count(p[i])) P.erase(p[i]);
        if (P.count(q[i])) P.erase(q[i]);
      }
      ++i;
      if (P.empty()) break;
    }
    b.push_back(i);
    s = i;
  }
  int B = b.size();
  if (B >= k) {
    string ans(n, '?');
    int pre = 0;
    for (int i = 0; i < (int)(B); ++i) {
      char c = 'a' + min(i, k - 1);
      for (int j = pre; j < b[i]; ++j) ans[p[j]] = c;
      pre = b[i];
    }
    cout << "YES\n";
    cout << ans << "\n";
  } else
    cout << "NO\n";
  return 0;
}

```