#include <bits/stdc++.h>
using namespace std;
long long N, K;
long long a[200005];
long long b[200005];
long long pos[200005];
char c[200005];
char d[200005];
long long f(long long x) { return x * (x + 1) / 2; }
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> N >> K;
  for (long long i = 1; i <= N; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (long long i = 1; i <= N; i++) {
    cin >> b[i];
    b[i] = pos[b[i]];
  }
  vector<long long> gpo;
  long long l = 1;
  while (l <= N) {
    long long nl = l;
    long long nr = b[l];
    long long sum = b[l];
    long long r = l;
    while (r - l != nr - nl or sum != f(nr) - f(nl - 1)) {
      r++;
      sum += b[r];
      nr = max(nr, b[r]);
    }
    gpo.push_back(l);
    l = r + 1;
  }
  while (gpo.size() > K) gpo.pop_back();
  if (gpo.size() < K) {
    cout << "NO\n";
    exit(0);
  }
  gpo.push_back(N + 1);
  for (long long i = 0; i < gpo.size() - 1; i++) {
    long long l = gpo[i];
    long long r = gpo[i + 1] - 1;
    while (l <= r) c[l++] = 'a' + i;
  }
  for (long long i = 1; i <= N; i++) d[a[i]] = c[i];
  cout << "YES\n";
  for (long long i = 1; i <= N; i++) cout << d[i];
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long N, K;
long long a[200005];
long long b[200005];
long long pos[200005];
char c[200005];
char d[200005];
long long f(long long x) { return x * (x + 1) / 2; }
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> N >> K;
  for (long long i = 1; i <= N; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (long long i = 1; i <= N; i++) {
    cin >> b[i];
    b[i] = pos[b[i]];
  }
  vector<long long> gpo;
  long long l = 1;
  while (l <= N) {
    long long nl = l;
    long long nr = b[l];
    long long sum = b[l];
    long long r = l;
    while (r - l != nr - nl or sum != f(nr) - f(nl - 1)) {
      r++;
      sum += b[r];
      nr = max(nr, b[r]);
    }
    gpo.push_back(l);
    l = r + 1;
  }
  while (gpo.size() > K) gpo.pop_back();
  if (gpo.size() < K) {
    cout << "NO\n";
    exit(0);
  }
  gpo.push_back(N + 1);
  for (long long i = 0; i < gpo.size() - 1; i++) {
    long long l = gpo[i];
    long long r = gpo[i + 1] - 1;
    while (l <= r) c[l++] = 'a' + i;
  }
  for (long long i = 1; i <= N; i++) d[a[i]] = c[i];
  cout << "YES\n";
  for (long long i = 1; i <= N; i++) cout << d[i];
  return 0;
}

```