#include <bits/stdc++.h>
using namespace std;
const long long logN = 20;
const long long M = 1000000007;
const long long INF = 1e12;
const long long N = 200005;
long long A[N], B[N];
set<long long> P, Q;
vector<long long> pos;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> A[i];
  for (long long i = 0; i < n; i++) cin >> B[i];
  for (long long i = 0; i < n; i++) {
    if (Q.count(A[i])) {
      Q.erase(A[i]);
    } else {
      P.insert(A[i]);
    }
    if (P.count(B[i])) {
      P.erase(B[i]);
    } else {
      Q.insert(B[i]);
    }
    if (P.empty() and Q.empty()) {
      pos.push_back(i);
    }
  }
  if (pos.size() < m) {
    cout << "NO"
         << "\n";
    return 0;
  }
  cout << "YES"
       << "\n";
  string s(n, ' ');
  long long l = 0;
  for (long long i = 0; i < pos.size(); i++) {
    long long r = pos[i];
    for (long long j = l; j < r + 1; j++) s[A[j] - 1] = ('a' + min(i, 25LL));
    l = r + 1;
  }
  cout << s << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP 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 long long logN = 20;
const long long M = 1000000007;
const long long INF = 1e12;
const long long N = 200005;
long long A[N], B[N];
set<long long> P, Q;
vector<long long> pos;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> A[i];
  for (long long i = 0; i < n; i++) cin >> B[i];
  for (long long i = 0; i < n; i++) {
    if (Q.count(A[i])) {
      Q.erase(A[i]);
    } else {
      P.insert(A[i]);
    }
    if (P.count(B[i])) {
      P.erase(B[i]);
    } else {
      Q.insert(B[i]);
    }
    if (P.empty() and Q.empty()) {
      pos.push_back(i);
    }
  }
  if (pos.size() < m) {
    cout << "NO"
         << "\n";
    return 0;
  }
  cout << "YES"
       << "\n";
  string s(n, ' ');
  long long l = 0;
  for (long long i = 0; i < pos.size(); i++) {
    long long r = pos[i];
    for (long long j = l; j < r + 1; j++) s[A[j] - 1] = ('a' + min(i, 25LL));
    l = r + 1;
  }
  cout << s << "\n";
  return 0;
}

```