#include <bits/stdc++.h>
using namespace std;
const long long MXN = 2e5 + 10;
long long n, k, ans, x = 1;
long long A[MXN], B[MXN], Comp[MXN], Ans[MXN];
vector<long long> adj[MXN], adt[MXN], Top;
bool mark[MXN];
char ch = 'a';
void dfs(long long u) {
  mark[u] = 1;
  for (auto v : adj[u]) {
    if (!mark[v]) dfs(v);
  }
  Top.push_back(u);
}
void sfd(long long u) {
  Comp[u] = x;
  for (auto v : adt[u]) {
    if (!Comp[v]) sfd(v);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1; i <= n; i++) cin >> B[i];
  for (int i = 1; i < n; i++) {
    long long u = A[i], v = A[i + 1];
    adj[u].push_back(v), adt[v].push_back(u);
    u = B[i], v = B[i + 1];
    adj[u].push_back(v), adt[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) dfs(i);
  }
  reverse(Top.begin(), Top.end());
  for (auto u : Top) {
    if (!Comp[u]) sfd(u), Ans[x] = (char)ch, x++, ch = min(ch + 1, (int)'z');
  }
  if (x - 1 < k) return cout << "NO", 0;
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) cout << (char)Ans[Comp[i]];
  return 0;
}

### Prompt

Create a solution in Cpp for 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 MXN = 2e5 + 10;
long long n, k, ans, x = 1;
long long A[MXN], B[MXN], Comp[MXN], Ans[MXN];
vector<long long> adj[MXN], adt[MXN], Top;
bool mark[MXN];
char ch = 'a';
void dfs(long long u) {
  mark[u] = 1;
  for (auto v : adj[u]) {
    if (!mark[v]) dfs(v);
  }
  Top.push_back(u);
}
void sfd(long long u) {
  Comp[u] = x;
  for (auto v : adt[u]) {
    if (!Comp[v]) sfd(v);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1; i <= n; i++) cin >> B[i];
  for (int i = 1; i < n; i++) {
    long long u = A[i], v = A[i + 1];
    adj[u].push_back(v), adt[v].push_back(u);
    u = B[i], v = B[i + 1];
    adj[u].push_back(v), adt[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) dfs(i);
  }
  reverse(Top.begin(), Top.end());
  for (auto u : Top) {
    if (!Comp[u]) sfd(u), Ans[x] = (char)ch, x++, ch = min(ch + 1, (int)'z');
  }
  if (x - 1 < k) return cout << "NO", 0;
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) cout << (char)Ans[Comp[i]];
  return 0;
}

```