#include <bits/stdc++.h>
using namespace std;
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int i, j, u, v, x, y, p, t, m, k, l, r, w, n;
  cin >> n >> k;
  long long int arr1[n], arr2[n];
  map<long long int, long long int> mpp1, mpp2;
  for (i = 0; i < n; i++) {
    cin >> arr1[i];
    mpp1[arr1[i]] = i;
  }
  for (i = 0; i < n; i++) {
    cin >> arr2[i];
    mpp2[arr2[i]] = i;
  }
  char ans[n + 1];
  for (i = 1; i <= n; i++) {
    ans[i] = 'a';
  }
  char ch = 'z';
  long long int ind1 = n - 1, ind2 = n - 1;
  while (ind1 >= 0 && ind2 >= 0 && ch > 'a') {
    long long int temp1 = mpp2[arr1[ind1]], temp2 = mpp1[arr2[ind2]];
    while (ind1 > temp2 || ind2 > temp1) {
      ans[arr1[ind1]] = ch;
      ans[arr2[ind2]] = ch;
      if (ind1 > temp2) ind1--;
      if (ind2 > temp1) ind2--;
      if (ind1 > 0) {
        temp2 = min(temp2, mpp1[arr2[ind2]]);
        temp1 = min(temp1, mpp2[arr1[ind1]]);
      }
    }
    ans[arr1[ind1]] = ch;
    ans[arr2[ind2]] = ch;
    ind1--;
    ind2--;
    ch--;
  }
  set<char> s;
  for (i = 1; i <= n; i++) {
    s.insert(ans[i]);
  }
  if (s.size() < k) {
    cout << "NO";
    return 0;
  }
  cout << "YES"
       << "\n";
  for (i = 1; i <= n; i++) {
    cout << ans[i];
  }
}

### Prompt

Your challenge is to write 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;
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int i, j, u, v, x, y, p, t, m, k, l, r, w, n;
  cin >> n >> k;
  long long int arr1[n], arr2[n];
  map<long long int, long long int> mpp1, mpp2;
  for (i = 0; i < n; i++) {
    cin >> arr1[i];
    mpp1[arr1[i]] = i;
  }
  for (i = 0; i < n; i++) {
    cin >> arr2[i];
    mpp2[arr2[i]] = i;
  }
  char ans[n + 1];
  for (i = 1; i <= n; i++) {
    ans[i] = 'a';
  }
  char ch = 'z';
  long long int ind1 = n - 1, ind2 = n - 1;
  while (ind1 >= 0 && ind2 >= 0 && ch > 'a') {
    long long int temp1 = mpp2[arr1[ind1]], temp2 = mpp1[arr2[ind2]];
    while (ind1 > temp2 || ind2 > temp1) {
      ans[arr1[ind1]] = ch;
      ans[arr2[ind2]] = ch;
      if (ind1 > temp2) ind1--;
      if (ind2 > temp1) ind2--;
      if (ind1 > 0) {
        temp2 = min(temp2, mpp1[arr2[ind2]]);
        temp1 = min(temp1, mpp2[arr1[ind1]]);
      }
    }
    ans[arr1[ind1]] = ch;
    ans[arr2[ind2]] = ch;
    ind1--;
    ind2--;
    ch--;
  }
  set<char> s;
  for (i = 1; i <= n; i++) {
    s.insert(ans[i]);
  }
  if (s.size() < k) {
    cout << "NO";
    return 0;
  }
  cout << "YES"
       << "\n";
  for (i = 1; i <= n; i++) {
    cout << ans[i];
  }
}

```