#include <bits/stdc++.h>
using namespace std;
const double pi = 3.1415926535897932384626433832795;
const int inf = 1e9;
const long long inf2 = 1e18;
const int mod = 1e9 + 7;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n), pos(n, 0), q(n), r(n, 0), l(n, 0);
  iota(r.begin(), r.end(), 0);
  iota(l.begin(), l.end(), 0);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    p[i]--;
    pos[p[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    cin >> q[i];
    q[i]--;
  }
  for (int i = 0; i < n - 1; i++) {
    if (pos[q[i]] > pos[q[i + 1]]) {
      l[pos[q[i]]] = pos[q[i + 1]];
    }
  }
  int x = 0, y = 0, curl = n;
  for (int i = n - 1; i >= 0; i--) {
    if (curl > i) {
      y = 0;
    }
    curl = min(curl, l[i]);
    y = max(y, i);
    r[i] = max(r[i], y);
  }
  vector<int> ans(n, 0);
  k--;
  int lst = 0;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j <= r[i]) {
      ans[p[j]] = x;
      lst = x;
      j++;
    }
    i = j - 1;
    if (x < k) x++;
  }
  if (lst < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) {
    cout << (char)(ans[i] + 'a');
  }
  return;
}
int main() {
  boost();
  int tc = 1;
  while (tc--) solve();
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const double pi = 3.1415926535897932384626433832795;
const int inf = 1e9;
const long long inf2 = 1e18;
const int mod = 1e9 + 7;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n), pos(n, 0), q(n), r(n, 0), l(n, 0);
  iota(r.begin(), r.end(), 0);
  iota(l.begin(), l.end(), 0);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    p[i]--;
    pos[p[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    cin >> q[i];
    q[i]--;
  }
  for (int i = 0; i < n - 1; i++) {
    if (pos[q[i]] > pos[q[i + 1]]) {
      l[pos[q[i]]] = pos[q[i + 1]];
    }
  }
  int x = 0, y = 0, curl = n;
  for (int i = n - 1; i >= 0; i--) {
    if (curl > i) {
      y = 0;
    }
    curl = min(curl, l[i]);
    y = max(y, i);
    r[i] = max(r[i], y);
  }
  vector<int> ans(n, 0);
  k--;
  int lst = 0;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j <= r[i]) {
      ans[p[j]] = x;
      lst = x;
      j++;
    }
    i = j - 1;
    if (x < k) x++;
  }
  if (lst < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) {
    cout << (char)(ans[i] + 'a');
  }
  return;
}
int main() {
  boost();
  int tc = 1;
  while (tc--) solve();
  return 0;
}

```