#include <bits/stdc++.h>
using namespace std;
int mark[210000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    memset(mark, 0, sizeof(mark));
    int arr[210000], brr[210000];
    int n, k;
    cin >> n >> k;
    string s = "";
    for (int i = 0; i < n; i++) cin >> arr[i];
    for (int i = 0; i < n; i++) cin >> brr[i];
    char ch = 'a';
    for (int i = 0; i < n; i++) {
      if (arr[i] == brr[i]) {
        s += ch;
        if (ch < 'z' && k > 0) {
          ch++;
        }
      } else {
        int j = i;
        mark[brr[j]] = 1;
        for (int k = i; k <= j && k < n; k++) {
          if (mark[arr[k]] == 0) {
            while (brr[j] != arr[k] && j < n) {
              mark[brr[j]] = 1;
              j++;
            }
            mark[brr[j]] = 1;
          }
          s += ch;
        }
        i = j;
        if (ch < 'z' && k > 0) {
          ch++;
        }
      }
    }
    ch = 'a';
    for (int i = 0; i < n; i++)
      if (s[i] > ch) ch = s[i];
    ch++;
    if (k <= (ch - 'a')) {
      string s1 = s;
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        s1[arr[i] - 1] = s[i];
      }
      cout << s1 << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}

### 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 mark[210000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    memset(mark, 0, sizeof(mark));
    int arr[210000], brr[210000];
    int n, k;
    cin >> n >> k;
    string s = "";
    for (int i = 0; i < n; i++) cin >> arr[i];
    for (int i = 0; i < n; i++) cin >> brr[i];
    char ch = 'a';
    for (int i = 0; i < n; i++) {
      if (arr[i] == brr[i]) {
        s += ch;
        if (ch < 'z' && k > 0) {
          ch++;
        }
      } else {
        int j = i;
        mark[brr[j]] = 1;
        for (int k = i; k <= j && k < n; k++) {
          if (mark[arr[k]] == 0) {
            while (brr[j] != arr[k] && j < n) {
              mark[brr[j]] = 1;
              j++;
            }
            mark[brr[j]] = 1;
          }
          s += ch;
        }
        i = j;
        if (ch < 'z' && k > 0) {
          ch++;
        }
      }
    }
    ch = 'a';
    for (int i = 0; i < n; i++)
      if (s[i] > ch) ch = s[i];
    ch++;
    if (k <= (ch - 'a')) {
      string s1 = s;
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        s1[arr[i] - 1] = s[i];
      }
      cout << s1 << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}

```