#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
vector<vector<long long> > g(N), rg(N);
vector<long long> order, component;
bool vis[N];
char ans[N];
void dfs1(long long src) {
  vis[src] = 1;
  for (auto i : g[src]) {
    if (!vis[i]) {
      dfs1(i);
    }
  }
  order.push_back(src);
}
void dfs2(long long src) {
  vis[src] = 1;
  component.push_back(src);
  for (auto i : rg[src]) {
    if (!vis[i]) {
      dfs2(i);
    }
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    g[a[i]].push_back(a[i + 1]);
    rg[a[i + 1]].push_back(a[i]);
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    g[a[i]].push_back(a[i + 1]);
    rg[a[i + 1]].push_back(a[i]);
  }
  memset(vis, 0, sizeof(vis));
  for (long long i = 1; i < n + 1; i++) {
    if (!vis[i]) {
      dfs1(i);
    }
  }
  memset(vis, 0, sizeof(vis));
  long long z = 0;
  string ff = "abcdefghijklmnopqrstuvwxyz";
  for (long long i = 0; i < n; i++) {
    long long v = order[n - 1 - i];
    if (!vis[v]) {
      dfs2(v);
      for (auto j : component) {
        ans[j] = ff[min(z, k - 1)];
      }
      z++;
      component.clear();
    }
  }
  if (z < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (long long i = 1; i < n + 1; i++) {
    cout << ans[i];
  }
  cout << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tests = 1;
  while (tests--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const long long N = 2e5 + 5;
vector<vector<long long> > g(N), rg(N);
vector<long long> order, component;
bool vis[N];
char ans[N];
void dfs1(long long src) {
  vis[src] = 1;
  for (auto i : g[src]) {
    if (!vis[i]) {
      dfs1(i);
    }
  }
  order.push_back(src);
}
void dfs2(long long src) {
  vis[src] = 1;
  component.push_back(src);
  for (auto i : rg[src]) {
    if (!vis[i]) {
      dfs2(i);
    }
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    g[a[i]].push_back(a[i + 1]);
    rg[a[i + 1]].push_back(a[i]);
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    g[a[i]].push_back(a[i + 1]);
    rg[a[i + 1]].push_back(a[i]);
  }
  memset(vis, 0, sizeof(vis));
  for (long long i = 1; i < n + 1; i++) {
    if (!vis[i]) {
      dfs1(i);
    }
  }
  memset(vis, 0, sizeof(vis));
  long long z = 0;
  string ff = "abcdefghijklmnopqrstuvwxyz";
  for (long long i = 0; i < n; i++) {
    long long v = order[n - 1 - i];
    if (!vis[v]) {
      dfs2(v);
      for (auto j : component) {
        ans[j] = ff[min(z, k - 1)];
      }
      z++;
      component.clear();
    }
  }
  if (z < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (long long i = 1; i < n + 1; i++) {
    cout << ans[i];
  }
  cout << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tests = 1;
  while (tests--) {
    solve();
  }
  return 0;
}

```