#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
int p[MAXN], q[MAXN], ans[MAXN];
vector<int> adj[MAXN], radj[MAXN];
bool visited[MAXN];
void dfs(int now, vector<int> adj[], vector<int> &urut) {
  visited[now] = true;
  for (auto nxt : adj[now]) {
    if (!visited[nxt]) {
      dfs(nxt, adj, urut);
    }
  }
  urut.push_back(now);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (i > 1) {
      adj[p[i - 1]].push_back(p[i]);
      radj[p[i]].push_back(p[i - 1]);
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> q[i];
    if (i > 1) {
      adj[q[i - 1]].push_back(q[i]);
      radj[q[i]].push_back(q[i - 1]);
    }
  }
  vector<int> order;
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) dfs(i, adj, order);
  }
  reverse(order.begin(), order.end());
  memset(visited, false, sizeof(visited));
  int cnt = 0, cntSCC = 0;
  for (auto x : order) {
    if (!visited[x]) {
      vector<int> scc;
      dfs(x, radj, scc);
      for (auto isi : scc) ans[isi] = cnt;
      if (cnt < 25) cnt++;
      cntSCC++;
    }
  }
  if (cntSCC < k)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << char((int)'a' + ans[i]);
    cout << '\n';
  }
  return 0;
}

### Prompt

Generate 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 int MAXN = 200005;
int p[MAXN], q[MAXN], ans[MAXN];
vector<int> adj[MAXN], radj[MAXN];
bool visited[MAXN];
void dfs(int now, vector<int> adj[], vector<int> &urut) {
  visited[now] = true;
  for (auto nxt : adj[now]) {
    if (!visited[nxt]) {
      dfs(nxt, adj, urut);
    }
  }
  urut.push_back(now);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (i > 1) {
      adj[p[i - 1]].push_back(p[i]);
      radj[p[i]].push_back(p[i - 1]);
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> q[i];
    if (i > 1) {
      adj[q[i - 1]].push_back(q[i]);
      radj[q[i]].push_back(q[i - 1]);
    }
  }
  vector<int> order;
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) dfs(i, adj, order);
  }
  reverse(order.begin(), order.end());
  memset(visited, false, sizeof(visited));
  int cnt = 0, cntSCC = 0;
  for (auto x : order) {
    if (!visited[x]) {
      vector<int> scc;
      dfs(x, radj, scc);
      for (auto isi : scc) ans[isi] = cnt;
      if (cnt < 25) cnt++;
      cntSCC++;
    }
  }
  if (cntSCC < k)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << char((int)'a' + ans[i]);
    cout << '\n';
  }
  return 0;
}

```