#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(239);
vector<int> f(vector<int> a) {
  int n = ((int)((a).size()));
  vector<int> b(n);
  for (int i = 0; i < n; i++) b[a[i]] = i;
  return b;
}
int solve() {
  int n;
  if (!(cin >> n)) return 1;
  int k;
  cin >> k;
  vector<int> p(n), q(n);
  for (int &i : p) cin >> i, i--;
  for (int &i : q) cin >> i, i--;
  p = f(p);
  q = f(q);
  vector<int> qq(n);
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    qq[i] = p[q[i]];
    int l = min(p[i], q[i]);
    int r = max(p[i], q[i]);
    v.push_back({l, -1});
    v.push_back({r, 1});
  }
  sort(v.begin(), v.end());
  int beg = v[0].first;
  int bal = 0;
  vector<int> eq(n, 0);
  for (auto it : v) {
    int pos = it.first;
    int t = it.second;
    bal -= t;
    if (bal == 0) {
      for (int i = beg + 1; i <= pos; i++) eq[i] = 1;
    } else if (t == -1 && bal == 1) {
      beg = pos;
    }
  }
  int x = 0;
  for (int i : eq)
    if (!i) x++;
  if (x < k) {
    cout << "NO" << endl;
    return 1;
  }
  string s = "";
  int cur = -1;
  for (int i = 0; i < n; i++) {
    if (!eq[i]) cur = min(cur + 1, k - 1);
    s.push_back('a' + cur);
  }
  vector<int> revp = f(p);
  string t = "";
  for (int i = 0; i < n; i++) t.push_back('a');
  for (int i = 0; i < n; i++) t[revp[i]] = s[i];
  cout << "YES\n" << t << endl;
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TET = 1e9;
  while (TET--) {
    if (solve()) break;
  }
}

### Prompt

Generate 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 rnd(239);
vector<int> f(vector<int> a) {
  int n = ((int)((a).size()));
  vector<int> b(n);
  for (int i = 0; i < n; i++) b[a[i]] = i;
  return b;
}
int solve() {
  int n;
  if (!(cin >> n)) return 1;
  int k;
  cin >> k;
  vector<int> p(n), q(n);
  for (int &i : p) cin >> i, i--;
  for (int &i : q) cin >> i, i--;
  p = f(p);
  q = f(q);
  vector<int> qq(n);
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    qq[i] = p[q[i]];
    int l = min(p[i], q[i]);
    int r = max(p[i], q[i]);
    v.push_back({l, -1});
    v.push_back({r, 1});
  }
  sort(v.begin(), v.end());
  int beg = v[0].first;
  int bal = 0;
  vector<int> eq(n, 0);
  for (auto it : v) {
    int pos = it.first;
    int t = it.second;
    bal -= t;
    if (bal == 0) {
      for (int i = beg + 1; i <= pos; i++) eq[i] = 1;
    } else if (t == -1 && bal == 1) {
      beg = pos;
    }
  }
  int x = 0;
  for (int i : eq)
    if (!i) x++;
  if (x < k) {
    cout << "NO" << endl;
    return 1;
  }
  string s = "";
  int cur = -1;
  for (int i = 0; i < n; i++) {
    if (!eq[i]) cur = min(cur + 1, k - 1);
    s.push_back('a' + cur);
  }
  vector<int> revp = f(p);
  string t = "";
  for (int i = 0; i < n; i++) t.push_back('a');
  for (int i = 0; i < n; i++) t[revp[i]] = s[i];
  cout << "YES\n" << t << endl;
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TET = 1e9;
  while (TET--) {
    if (solve()) break;
  }
}

```