#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 2e5 + 7;
bool vis1[MAX_N], vis2[MAX_N];
int count(int pos) {
  if (vis1[pos] == vis2[pos]) return 0;
  return 1;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  string s = "";
  cin >> n >> k;
  int p1[n], p2[n];
  for (int i = 0; i < n; i++) cin >> p1[i];
  for (int i = 0; i < n; i++) cin >> p2[i];
  for (int i = 0; i < n; i++) s += "a";
  int cnt = 0;
  memset(vis1, 0, sizeof(vis1));
  memset(vis2, 0, sizeof(vis2));
  int cur = 1;
  bool used = 0;
  for (int i = 0; i < n; i++) {
    if (!used) used = 1;
    s[p1[i] - 1] = 'a' + cur - 1;
    cnt -= count(p1[i]);
    cnt -= count(p2[i]);
    vis1[p1[i]] = vis2[p2[i]] = 1;
    cnt += count(p1[i]);
    cnt += count(p2[i]);
    if (cnt == 0 && cur < k) cur++, used = 0;
  }
  if (cur != k || !used) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  cout << s << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 MAX_N = 2e5 + 7;
bool vis1[MAX_N], vis2[MAX_N];
int count(int pos) {
  if (vis1[pos] == vis2[pos]) return 0;
  return 1;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  string s = "";
  cin >> n >> k;
  int p1[n], p2[n];
  for (int i = 0; i < n; i++) cin >> p1[i];
  for (int i = 0; i < n; i++) cin >> p2[i];
  for (int i = 0; i < n; i++) s += "a";
  int cnt = 0;
  memset(vis1, 0, sizeof(vis1));
  memset(vis2, 0, sizeof(vis2));
  int cur = 1;
  bool used = 0;
  for (int i = 0; i < n; i++) {
    if (!used) used = 1;
    s[p1[i] - 1] = 'a' + cur - 1;
    cnt -= count(p1[i]);
    cnt -= count(p2[i]);
    vis1[p1[i]] = vis2[p2[i]] = 1;
    cnt += count(p1[i]);
    cnt += count(p2[i]);
    if (cnt == 0 && cur < k) cur++, used = 0;
  }
  if (cur != k || !used) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  cout << s << endl;
  return 0;
}

```