#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)7 + (int)1e9;
const int N = 1000005;
const double PI = 4 * atan(1);
const double eps = 1e-10;
const long long oo = 1e10;
const int K = 26;
int n, k;
int p[N];
int q[N];
int ans[N];
vector<long long> adj[N];
vector<long long> adjt[N];
bool vis[N];
vector<long long> v;
vector<long long> order;
void dfs(int src) {
  if (vis[src]) return;
  vis[src] = 1;
  for (auto t : adj[src]) {
    if (vis[t]) continue;
    dfs(t);
  }
  order.push_back(src);
}
void label(int src) {
  if (vis[src]) return;
  v.push_back(src);
  vis[src] = 1;
  for (auto t : adjt[src]) {
    if (vis[t]) continue;
    label(t);
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    if (i > 0) {
      adj[p[i - 1]].push_back(p[i]);
      adjt[p[i]].push_back(p[i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    q[i]--;
    if (i > 0) {
      adj[q[i - 1]].push_back(q[i]);
      adjt[q[i]].push_back(q[i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  memset(vis, 0, sizeof(vis));
  int cur = 0;
  for (int i = 0; i < n; i++) {
    int node = order[n - 1 - i];
    if (vis[node]) continue;
    v.clear();
    label(node);
    for (auto t : v) {
      ans[t] = (cur < 25 ? cur : 25);
    }
    cur++;
  }
  if (cur < k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << char(ans[i] + 'a');
  }
  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 int MOD = (int)7 + (int)1e9;
const int N = 1000005;
const double PI = 4 * atan(1);
const double eps = 1e-10;
const long long oo = 1e10;
const int K = 26;
int n, k;
int p[N];
int q[N];
int ans[N];
vector<long long> adj[N];
vector<long long> adjt[N];
bool vis[N];
vector<long long> v;
vector<long long> order;
void dfs(int src) {
  if (vis[src]) return;
  vis[src] = 1;
  for (auto t : adj[src]) {
    if (vis[t]) continue;
    dfs(t);
  }
  order.push_back(src);
}
void label(int src) {
  if (vis[src]) return;
  v.push_back(src);
  vis[src] = 1;
  for (auto t : adjt[src]) {
    if (vis[t]) continue;
    label(t);
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    if (i > 0) {
      adj[p[i - 1]].push_back(p[i]);
      adjt[p[i]].push_back(p[i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    q[i]--;
    if (i > 0) {
      adj[q[i - 1]].push_back(q[i]);
      adjt[q[i]].push_back(q[i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  memset(vis, 0, sizeof(vis));
  int cur = 0;
  for (int i = 0; i < n; i++) {
    int node = order[n - 1 - i];
    if (vis[node]) continue;
    v.clear();
    label(node);
    for (auto t : v) {
      ans[t] = (cur < 25 ? cur : 25);
    }
    cur++;
  }
  if (cur < k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << char(ans[i] + 'a');
  }
  return 0;
}

```