#include <bits/stdc++.h>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(322179);
}
signed main() {
  doRoutine();
  int n, k;
  cin >> n >> k;
  vector<int> p(n), q(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    --p[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> q[i];
    --q[i];
  }
  vector<int> pos_in_p(n);
  for (int i = 0; i < n; ++i) {
    pos_in_p[p[i]] = i;
  }
  vector<int> trans(n);
  for (int i = 0; i + 1 < n; ++i) {
    int from = pos_in_p[q[i]], to = pos_in_p[q[i + 1]];
    if (from > to) {
      ++trans[to];
      --trans[from];
    }
  }
  for (int i = 1; i < n - 1; ++i) {
    trans[i] += trans[i - 1];
  }
  vector<int> color(n, -1);
  int cur_color = 0;
  for (int i = 0; i < n;) {
    int j = i + 1;
    color[i] = cur_color;
    while (j < n && trans[j - 1]) {
      color[j] = cur_color;
      ++j;
    }
    i = j;
    ++cur_color;
  }
  if (cur_color >= k) {
    for (int i = 0; i < n; ++i) {
      if (color[i] >= k - 1) {
        color[i] = k - 1;
      }
    }
    vector<int> ans(n);
    for (int i = 0; i < n; ++i) {
      ans[p[i]] = color[i];
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; ++i) {
      cout << (char)('a' + ans[i]);
    }
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(322179);
}
signed main() {
  doRoutine();
  int n, k;
  cin >> n >> k;
  vector<int> p(n), q(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    --p[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> q[i];
    --q[i];
  }
  vector<int> pos_in_p(n);
  for (int i = 0; i < n; ++i) {
    pos_in_p[p[i]] = i;
  }
  vector<int> trans(n);
  for (int i = 0; i + 1 < n; ++i) {
    int from = pos_in_p[q[i]], to = pos_in_p[q[i + 1]];
    if (from > to) {
      ++trans[to];
      --trans[from];
    }
  }
  for (int i = 1; i < n - 1; ++i) {
    trans[i] += trans[i - 1];
  }
  vector<int> color(n, -1);
  int cur_color = 0;
  for (int i = 0; i < n;) {
    int j = i + 1;
    color[i] = cur_color;
    while (j < n && trans[j - 1]) {
      color[j] = cur_color;
      ++j;
    }
    i = j;
    ++cur_color;
  }
  if (cur_color >= k) {
    for (int i = 0; i < n; ++i) {
      if (color[i] >= k - 1) {
        color[i] = k - 1;
      }
    }
    vector<int> ans(n);
    for (int i = 0; i < n; ++i) {
      ans[p[i]] = color[i];
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; ++i) {
      cout << (char)('a' + ans[i]);
    }
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

```