#include <bits/stdc++.h>
using namespace std;
constexpr int SIZE = (int)1e6;
int n, k, from, to;
vector<int> g[SIZE], h[SIZE];
vector<int> ord;
vector<bool> check;
string ans;
int component[SIZE];
int comp;
void dfs1(int v) {
  check[v] = true;
  for (int i : g[v]) {
    if (!check[i]) {
      dfs1(i);
    }
  }
  ord.push_back(v);
}
void dfs2(int v) {
  check[v] = true;
  component[v] = comp;
  for (int i : h[v]) {
    if (!check[i]) {
      dfs2(i);
    }
  }
}
int main() {
  cin >> n >> k;
  for (int j = 0; j < 2; j++) {
    cin >> from;
    from--;
    for (int i = 1; i < n; i++) {
      cin >> to;
      to--;
      g[from].push_back(to);
      h[to].push_back(from);
      from = to;
    }
  }
  check.resize(n, false);
  for (int i = 0; i < n; i++) {
    if (!check[i]) {
      dfs1(i);
    }
  }
  for (int i = 0; i < n; i++) {
    check[i] = false;
  }
  for (int i = ord.size() - 1; i >= 0; i--) {
    if (!check[ord[i]]) {
      dfs2(ord[i]);
      comp++;
    }
  }
  if (comp < k) {
    cout << "NO";
    return 0;
  }
  ans.resize(n);
  cout << "YES" << endl;
  for (int i = 0; i < ans.size(); i++) {
    if (k - 1 < component[i]) {
      ans[i] = char('a' + k - 1);
    } else {
      ans[i] = char('a' + component[i]);
    }
  }
  cout << ans;
  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;
constexpr int SIZE = (int)1e6;
int n, k, from, to;
vector<int> g[SIZE], h[SIZE];
vector<int> ord;
vector<bool> check;
string ans;
int component[SIZE];
int comp;
void dfs1(int v) {
  check[v] = true;
  for (int i : g[v]) {
    if (!check[i]) {
      dfs1(i);
    }
  }
  ord.push_back(v);
}
void dfs2(int v) {
  check[v] = true;
  component[v] = comp;
  for (int i : h[v]) {
    if (!check[i]) {
      dfs2(i);
    }
  }
}
int main() {
  cin >> n >> k;
  for (int j = 0; j < 2; j++) {
    cin >> from;
    from--;
    for (int i = 1; i < n; i++) {
      cin >> to;
      to--;
      g[from].push_back(to);
      h[to].push_back(from);
      from = to;
    }
  }
  check.resize(n, false);
  for (int i = 0; i < n; i++) {
    if (!check[i]) {
      dfs1(i);
    }
  }
  for (int i = 0; i < n; i++) {
    check[i] = false;
  }
  for (int i = ord.size() - 1; i >= 0; i--) {
    if (!check[ord[i]]) {
      dfs2(ord[i]);
      comp++;
    }
  }
  if (comp < k) {
    cout << "NO";
    return 0;
  }
  ans.resize(n);
  cout << "YES" << endl;
  for (int i = 0; i < ans.size(); i++) {
    if (k - 1 < component[i]) {
      ans[i] = char('a' + k - 1);
    } else {
      ans[i] = char('a' + component[i]);
    }
  }
  cout << ans;
  return 0;
}

```