#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  int a[n], b[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
  }
  int ind1[n], ind2[n];
  for (int i = 0; i < n; i++) {
    ind1[a[i]] = i;
    ind2[b[i]] = i;
  }
  int c[n];
  for (int i = 0; i < n; i++) {
    c[i] = ind2[a[i]];
    if (i > 0) c[i] = max(c[i], c[i - 1]);
  }
  char ans[n];
  char d = 'a';
  int end = 0;
  int st = 0;
  for (int i = 0; i < n; i++) {
    int g = max(i, c[ind1[b[i]]]);
    end = max(end, g);
    if (i != end) continue;
    for (int j = st; j <= end; j++) ans[b[j]] = d;
    if (d != 'z') d++;
    st = i + 1;
    end = i + 1;
  }
  for (int i = st; i < n; i++) {
    ans[b[i]] = d;
  }
  if (d != 'z' && d - 'a' < k)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << ans[i];
  }
}

### Prompt

Your task is to create 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_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  int a[n], b[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
  }
  int ind1[n], ind2[n];
  for (int i = 0; i < n; i++) {
    ind1[a[i]] = i;
    ind2[b[i]] = i;
  }
  int c[n];
  for (int i = 0; i < n; i++) {
    c[i] = ind2[a[i]];
    if (i > 0) c[i] = max(c[i], c[i - 1]);
  }
  char ans[n];
  char d = 'a';
  int end = 0;
  int st = 0;
  for (int i = 0; i < n; i++) {
    int g = max(i, c[ind1[b[i]]]);
    end = max(end, g);
    if (i != end) continue;
    for (int j = st; j <= end; j++) ans[b[j]] = d;
    if (d != 'z') d++;
    st = i + 1;
    end = i + 1;
  }
  for (int i = st; i < n; i++) {
    ans[b[i]] = d;
  }
  if (d != 'z' && d - 'a' < k)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << ans[i];
  }
}

```