#include <bits/stdc++.h>
using namespace std;
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(n);
  vector<int> q(n);
  for (int i = 0; i < n; ++i) cin >> p[i];
  for (int i = 0; i < n; ++i) cin >> q[i];
  int i = 0;
  int currK = 0;
  string ans(n, 'a');
  set<int> couldBeEqual;
  while (i < n) {
    if (p[i] == q[i]) {
      if (i == 0) {
        ++currK;
        ans[p[0] - 1] = 'a';
      } else {
        if (currK < k) {
          ++currK;
          couldBeEqual.insert(p[i] - 1);
          ans[p[i] - 1] = char(ans[p[i - 1] - 1] + 1);
        } else
          ans[p[i] - 1] = char(ans[p[i - 1] - 1]);
      }
      ++i;
    } else {
      unordered_set<int> aux;
      aux.insert(p[i]);
      aux.insert(q[i]);
      int j = i + 1;
      while (not aux.empty()) {
        if (aux.count(p[j]) == 0)
          aux.insert(p[j]);
        else
          aux.erase(p[j]);
        if (aux.count(q[j]) == 0)
          aux.insert(q[j]);
        else
          aux.erase(q[j]);
        ++j;
      }
      char c;
      if (i == 0) {
        ++currK;
        c = 'a';
      } else {
        if (currK < k) {
          ++currK;
          c = char(ans[p[i - 1] - 1] + 1);
        } else
          c = ans[p[i - 1] - 1];
      }
      for (int m = i; m < j; ++m) ans[p[m] - 1] = c;
      i = j;
    }
    if (currK > 26) {
      if (currK - int(couldBeEqual.size()) > 26) {
        currK = -1;
        break;
      } else {
        const int moves = currK - 26;
        vector<int> idx(moves + 1);
        idx[0] = n;
        for (int i = 1; i <= moves; ++i) {
          set<int>::iterator it = --end(couldBeEqual);
          idx[i] = *it;
          couldBeEqual.erase(it);
        }
        for (int j = 1, gap = moves; j <= moves; ++j, --gap) {
          for (int m = idx[j]; m < idx[j - 1]; ++m)
            ans[p[m] - 1] = char(int(ans[p[m] - 1]) - gap);
        }
        currK = 26;
      }
    }
  }
  if (k <= currK)
    cout << "YES\n" << ans << "\n";
  else
    cout << "NO\n";
  return 0;
}

### 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;
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(n);
  vector<int> q(n);
  for (int i = 0; i < n; ++i) cin >> p[i];
  for (int i = 0; i < n; ++i) cin >> q[i];
  int i = 0;
  int currK = 0;
  string ans(n, 'a');
  set<int> couldBeEqual;
  while (i < n) {
    if (p[i] == q[i]) {
      if (i == 0) {
        ++currK;
        ans[p[0] - 1] = 'a';
      } else {
        if (currK < k) {
          ++currK;
          couldBeEqual.insert(p[i] - 1);
          ans[p[i] - 1] = char(ans[p[i - 1] - 1] + 1);
        } else
          ans[p[i] - 1] = char(ans[p[i - 1] - 1]);
      }
      ++i;
    } else {
      unordered_set<int> aux;
      aux.insert(p[i]);
      aux.insert(q[i]);
      int j = i + 1;
      while (not aux.empty()) {
        if (aux.count(p[j]) == 0)
          aux.insert(p[j]);
        else
          aux.erase(p[j]);
        if (aux.count(q[j]) == 0)
          aux.insert(q[j]);
        else
          aux.erase(q[j]);
        ++j;
      }
      char c;
      if (i == 0) {
        ++currK;
        c = 'a';
      } else {
        if (currK < k) {
          ++currK;
          c = char(ans[p[i - 1] - 1] + 1);
        } else
          c = ans[p[i - 1] - 1];
      }
      for (int m = i; m < j; ++m) ans[p[m] - 1] = c;
      i = j;
    }
    if (currK > 26) {
      if (currK - int(couldBeEqual.size()) > 26) {
        currK = -1;
        break;
      } else {
        const int moves = currK - 26;
        vector<int> idx(moves + 1);
        idx[0] = n;
        for (int i = 1; i <= moves; ++i) {
          set<int>::iterator it = --end(couldBeEqual);
          idx[i] = *it;
          couldBeEqual.erase(it);
        }
        for (int j = 1, gap = moves; j <= moves; ++j, --gap) {
          for (int m = idx[j]; m < idx[j - 1]; ++m)
            ans[p[m] - 1] = char(int(ans[p[m] - 1]) - gap);
        }
        currK = 26;
      }
    }
  }
  if (k <= currK)
    cout << "YES\n" << ans << "\n";
  else
    cout << "NO\n";
  return 0;
}

```