#include <bits/stdc++.h>
using namespace std;
const int N = 3e5;
const int oo = 2e9;
int n, k, p[2][N], ans[N];
set<int> st[2];
queue<int> q;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < 2; ++i)
    for (int j = 0; j < n; ++j) {
      cin >> p[i][j];
      --p[i][j];
    }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 2; ++j) {
      int v = (j + 1) % 2;
      if (st[v].find(p[j][i]) == st[v].end())
        st[j].insert(p[j][i]);
      else
        st[v].erase(p[j][i]);
    }
    if (st[0].empty() && st[1].empty()) q.push(i);
  }
  if (q.size() < k) {
    cout << "NO";
    return 0;
  }
  cout << "YES"
       << "\n";
  int l = 0, let = 0;
  while (!q.empty()) {
    int r = q.front();
    q.pop();
    for (int i = l; i <= r; ++i) ans[p[0][i]] = let;
    l = r + 1;
    let = min(let + 1, 25);
  }
  for (int i = 0; i < n; ++i) cout << char(ans[i] + 97);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = 3e5;
const int oo = 2e9;
int n, k, p[2][N], ans[N];
set<int> st[2];
queue<int> q;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < 2; ++i)
    for (int j = 0; j < n; ++j) {
      cin >> p[i][j];
      --p[i][j];
    }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 2; ++j) {
      int v = (j + 1) % 2;
      if (st[v].find(p[j][i]) == st[v].end())
        st[j].insert(p[j][i]);
      else
        st[v].erase(p[j][i]);
    }
    if (st[0].empty() && st[1].empty()) q.push(i);
  }
  if (q.size() < k) {
    cout << "NO";
    return 0;
  }
  cout << "YES"
       << "\n";
  int l = 0, let = 0;
  while (!q.empty()) {
    int r = q.front();
    q.pop();
    for (int i = l; i <= r; ++i) ans[p[0][i]] = let;
    l = r + 1;
    let = min(let + 1, 25);
  }
  for (int i = 0; i < n; ++i) cout << char(ans[i] + 97);
}

```