#include <bits/stdc++.h>
using namespace std;
int n, k;
void solve() {
  cin >> n >> k;
  vector<int> p(n + 1), q(n + 1), last(n + 1);
  vector<pair<int, int>> segs;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    last[p[i]] = i;
  }
  for (int j = 1; j <= n; ++j) {
    cin >> q[j];
    last[q[j]] = max(last[q[j]], j);
  }
  int pos = 1;
  int prev = 1;
  for (int i = 1; i <= n; ++i) {
    if (i > pos) {
      segs.push_back({prev, i - 1});
      prev = i;
    }
    pos = max(pos, last[p[i]]);
  }
  segs.push_back({prev, n});
  if ((int)segs.size() < k) {
    cout << "NO";
    return;
  } else {
    cout << "YES\n";
    vector<char> s(n + 1);
    int color = 0;
    for (auto &pa : segs) {
      for (int i = pa.first; i <= pa.second; ++i) {
        s[p[i]] = 'a' + color;
      }
      ++color;
      if (color > 25) color = 25;
    }
    for (int i = 1; i <= n; ++i) cout << s[i];
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int n, k;
void solve() {
  cin >> n >> k;
  vector<int> p(n + 1), q(n + 1), last(n + 1);
  vector<pair<int, int>> segs;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    last[p[i]] = i;
  }
  for (int j = 1; j <= n; ++j) {
    cin >> q[j];
    last[q[j]] = max(last[q[j]], j);
  }
  int pos = 1;
  int prev = 1;
  for (int i = 1; i <= n; ++i) {
    if (i > pos) {
      segs.push_back({prev, i - 1});
      prev = i;
    }
    pos = max(pos, last[p[i]]);
  }
  segs.push_back({prev, n});
  if ((int)segs.size() < k) {
    cout << "NO";
    return;
  } else {
    cout << "YES\n";
    vector<char> s(n + 1);
    int color = 0;
    for (auto &pa : segs) {
      for (int i = pa.first; i <= pa.second; ++i) {
        s[p[i]] = 'a' + color;
      }
      ++color;
      if (color > 25) color = 25;
    }
    for (int i = 1; i <= n; ++i) cout << s[i];
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

```