#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long base = 1e9 + 7;
const long long oo = 1e6 + 10;
const long long gh = 1e3 + 3;
const long long inf = 1e15 + 5;
long long n, m, k;
long long nextt[oo];
char res[oo];
long long a[oo], b[oo];
long long vta[oo];
long long vtb[oo];
string s, t;
void input() {
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    vta[a[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> b[i];
    vtb[b[i]] = i;
  }
}
char kitu;
void solve() {
  kitu = 'a';
  long long va = 1, vb = 1, na = 1, nb = 1;
  for (long long i = 1; i <= n; i++) {
    res[a[i]] = kitu;
    nb = max(nb, vtb[a[i]] + 1);
    while (vb < nb || va < na) {
      while (vb < nb) {
        na = max(na, vta[b[vb]] + 1);
        vb++;
      }
      while (va < na) {
        res[a[va]] = kitu;
        nb = max(nb, vtb[a[va]] + 1);
        va++;
      }
      i = max(i, va - 1);
    }
    if (kitu < 'z' || i == n) kitu++;
  }
}
void ouput() {
  if (kitu - 'a' < m) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (long long i = 1; i <= n; i++) cout << res[i];
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (ifstream("test.txt")) {
    freopen("test.txt", "r", stdin);
  } else if (ifstream(""
                      ".inp")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  long long t = 1;
  while (t--) {
    input();
    solve();
    ouput();
  }
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long base = 1e9 + 7;
const long long oo = 1e6 + 10;
const long long gh = 1e3 + 3;
const long long inf = 1e15 + 5;
long long n, m, k;
long long nextt[oo];
char res[oo];
long long a[oo], b[oo];
long long vta[oo];
long long vtb[oo];
string s, t;
void input() {
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    vta[a[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> b[i];
    vtb[b[i]] = i;
  }
}
char kitu;
void solve() {
  kitu = 'a';
  long long va = 1, vb = 1, na = 1, nb = 1;
  for (long long i = 1; i <= n; i++) {
    res[a[i]] = kitu;
    nb = max(nb, vtb[a[i]] + 1);
    while (vb < nb || va < na) {
      while (vb < nb) {
        na = max(na, vta[b[vb]] + 1);
        vb++;
      }
      while (va < na) {
        res[a[va]] = kitu;
        nb = max(nb, vtb[a[va]] + 1);
        va++;
      }
      i = max(i, va - 1);
    }
    if (kitu < 'z' || i == n) kitu++;
  }
}
void ouput() {
  if (kitu - 'a' < m) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (long long i = 1; i <= n; i++) cout << res[i];
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (ifstream("test.txt")) {
    freopen("test.txt", "r", stdin);
  } else if (ifstream(""
                      ".inp")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  long long t = 1;
  while (t--) {
    input();
    solve();
    ouput();
  }
}

```