#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k;
signed main() {
  n = in();
  k = in();
  vector<int> p1(n), p2(n);
  for (int i = 0; i < n; ++i) {
    p1[i] = in();
    --p1[i];
  }
  for (int i = 0; i < n; ++i) {
    p2[i] = in();
    --p2[i];
  }
  set<int> qq, pp;
  vector<int> aa;
  for (int i = 0; i < n; ++i) {
    if (pp.count(p1[i]))
      pp.erase(p1[i]);
    else
      qq.insert(p1[i]);
    if (qq.count(p2[i]))
      qq.erase(p2[i]);
    else
      pp.insert(p2[i]);
    if (qq.empty() && pp.empty()) aa.push_back(i);
  }
  if (int(aa.size()) < k)
    cout << "NO" << endl;
  else {
    string s(n, ' ');
    int l = 0;
    for (int t = 0; t < int(aa.size()); ++t) {
      int r = aa[t];
      char c = 'a' + min(t, 25);
      for (int i = l; i <= r; ++i) s[p1[i]] = c;
      l = r + 1;
    }
    cout << "YES" << endl << s << endl;
  }
  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;
inline int in() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k;
signed main() {
  n = in();
  k = in();
  vector<int> p1(n), p2(n);
  for (int i = 0; i < n; ++i) {
    p1[i] = in();
    --p1[i];
  }
  for (int i = 0; i < n; ++i) {
    p2[i] = in();
    --p2[i];
  }
  set<int> qq, pp;
  vector<int> aa;
  for (int i = 0; i < n; ++i) {
    if (pp.count(p1[i]))
      pp.erase(p1[i]);
    else
      qq.insert(p1[i]);
    if (qq.count(p2[i]))
      qq.erase(p2[i]);
    else
      pp.insert(p2[i]);
    if (qq.empty() && pp.empty()) aa.push_back(i);
  }
  if (int(aa.size()) < k)
    cout << "NO" << endl;
  else {
    string s(n, ' ');
    int l = 0;
    for (int t = 0; t < int(aa.size()); ++t) {
      int r = aa[t];
      char c = 'a' + min(t, 25);
      for (int i = l; i <= r; ++i) s[p1[i]] = c;
      l = r + 1;
    }
    cout << "YES" << endl << s << endl;
  }
  return 0;
}

```