#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
const long long mod = 1e9 + 7;
vector<vector<int>> adj(MAXN);
vector<vector<int>> radj(MAXN);
vector<bool> vis;
stack<int> order;
vector<int> component;
vector<int> letter(MAXN);
void dfs1(int u) {
  vis[u] = true;
  for (auto &it : adj[u])
    if (!vis[it]) dfs1(it);
  order.push(u);
}
void dfs2(int u) {
  vis[u] = true;
  for (auto &it : radj[u])
    if (!vis[it]) dfs2(it);
  component.push_back(u);
}
int scc(int n) {
  int count = 0;
  vis.resize(n + 1, false);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs1(i);
  fill(vis.begin(), vis.end(), false);
  while (!order.empty()) {
    int x = order.top();
    order.pop();
    if (!vis[x]) {
      dfs2(x);
      ;
      for (auto &it : component) letter[it] = count;
      component.clear();
      ++count;
    }
  }
  return count;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0) adj[a[i - 1]].push_back(a[i]), radj[a[i]].push_back(a[i - 1]);
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    if (i > 0) adj[b[i - 1]].push_back(b[i]), radj[b[i]].push_back(b[i - 1]);
  };
  ;
  int ans = scc(n);
  ;
  if (ans < k) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << (char)('a' + min(letter[i], 25));
    cout << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  for (int I = 1; I <= t; I++) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 int MAXN = 2e5 + 10;
const long long mod = 1e9 + 7;
vector<vector<int>> adj(MAXN);
vector<vector<int>> radj(MAXN);
vector<bool> vis;
stack<int> order;
vector<int> component;
vector<int> letter(MAXN);
void dfs1(int u) {
  vis[u] = true;
  for (auto &it : adj[u])
    if (!vis[it]) dfs1(it);
  order.push(u);
}
void dfs2(int u) {
  vis[u] = true;
  for (auto &it : radj[u])
    if (!vis[it]) dfs2(it);
  component.push_back(u);
}
int scc(int n) {
  int count = 0;
  vis.resize(n + 1, false);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs1(i);
  fill(vis.begin(), vis.end(), false);
  while (!order.empty()) {
    int x = order.top();
    order.pop();
    if (!vis[x]) {
      dfs2(x);
      ;
      for (auto &it : component) letter[it] = count;
      component.clear();
      ++count;
    }
  }
  return count;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0) adj[a[i - 1]].push_back(a[i]), radj[a[i]].push_back(a[i - 1]);
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    if (i > 0) adj[b[i - 1]].push_back(b[i]), radj[b[i]].push_back(b[i - 1]);
  };
  ;
  int ans = scc(n);
  ;
  if (ans < k) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << (char)('a' + min(letter[i], 25));
    cout << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  for (int I = 1; I <= t; I++) {
    solve();
  }
  return 0;
}

```