#include <bits/stdc++.h>
template <class... Args>
inline __attribute__((always_inline)) void read(Args&... args) {
  (std::cin >> ... >> args);
}
template <class... Args>
inline __attribute__((always_inline)) void write(Args&&... args) {
  (std::cout << ... << std::forward<Args>(args));
}
template <class... Args>
inline __attribute__((always_inline)) void writeln(Args&&... args) {
  write(std::forward<Args>(args)..., '\n');
}
inline __attribute__((always_inline)) void flush_write() {
  std::flush(std::cout);
}
class graph_t {
  using node_list_t = std::vector<int>;
  using holder_t = std::vector<node_list_t>;
  holder_t gr;

 public:
  graph_t(int n) : gr(n) {}
  void add_edge(int a, int b) { gr[b].push_back(a); }

 protected:
  void fill_comp(int v, std::vector<char>& used, std::vector<int>& comp) {
    used[v] = true;
    comp.push_back(v);
    for (int u : gr[v])
      if (!used[u]) fill_comp(u, used, comp);
  }

 public:
  std::vector<std::vector<int>> get_blocks(const std::vector<int>& order) {
    std::vector<std::vector<int>> blocks;
    std::vector<char> used(gr.size(), false);
    for (int v : order) {
      if (!used[v]) {
        blocks.emplace_back();
        fill_comp(v, used, blocks.back());
      }
    }
    return blocks;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::fixed(std::cout).precision(20);
  std::cin.exceptions(std::ios_base::failbit);
  std::cin.tie(nullptr);
  int n, k;
  read(n, k);
  std::vector<int> p1(n), p2(n);
  for (int& x : p1) read(x), --x;
  for (int& x : p2) read(x), --x;
  graph_t g(n);
  for (int i = 1; i < n; ++i)
    g.add_edge(p1[i - 1], p1[i]), g.add_edge(p2[i - 1], p2[i]);
  auto blocks = g.get_blocks(p1);
  if ((int)blocks.size() < k) {
    writeln("NO");
  } else {
    writeln("YES");
    std::string res(n, '-');
    int m = blocks.size();
    for (int i = 0; i < m; ++i)
      for (int x : blocks[i]) res[x] = 'a' + std::min(i, k - 1);
    writeln(res);
  }
  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>
template <class... Args>
inline __attribute__((always_inline)) void read(Args&... args) {
  (std::cin >> ... >> args);
}
template <class... Args>
inline __attribute__((always_inline)) void write(Args&&... args) {
  (std::cout << ... << std::forward<Args>(args));
}
template <class... Args>
inline __attribute__((always_inline)) void writeln(Args&&... args) {
  write(std::forward<Args>(args)..., '\n');
}
inline __attribute__((always_inline)) void flush_write() {
  std::flush(std::cout);
}
class graph_t {
  using node_list_t = std::vector<int>;
  using holder_t = std::vector<node_list_t>;
  holder_t gr;

 public:
  graph_t(int n) : gr(n) {}
  void add_edge(int a, int b) { gr[b].push_back(a); }

 protected:
  void fill_comp(int v, std::vector<char>& used, std::vector<int>& comp) {
    used[v] = true;
    comp.push_back(v);
    for (int u : gr[v])
      if (!used[u]) fill_comp(u, used, comp);
  }

 public:
  std::vector<std::vector<int>> get_blocks(const std::vector<int>& order) {
    std::vector<std::vector<int>> blocks;
    std::vector<char> used(gr.size(), false);
    for (int v : order) {
      if (!used[v]) {
        blocks.emplace_back();
        fill_comp(v, used, blocks.back());
      }
    }
    return blocks;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::fixed(std::cout).precision(20);
  std::cin.exceptions(std::ios_base::failbit);
  std::cin.tie(nullptr);
  int n, k;
  read(n, k);
  std::vector<int> p1(n), p2(n);
  for (int& x : p1) read(x), --x;
  for (int& x : p2) read(x), --x;
  graph_t g(n);
  for (int i = 1; i < n; ++i)
    g.add_edge(p1[i - 1], p1[i]), g.add_edge(p2[i - 1], p2[i]);
  auto blocks = g.get_blocks(p1);
  if ((int)blocks.size() < k) {
    writeln("NO");
  } else {
    writeln("YES");
    std::string res(n, '-');
    int m = blocks.size();
    for (int i = 0; i < m; ++i)
      for (int x : blocks[i]) res[x] = 'a' + std::min(i, k - 1);
    writeln(res);
  }
  return 0;
}

```