#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const int MOD = 1000000007;
char out[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(n), q(n), rnk(n), corr(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    rnk[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    q[i]--;
    corr[i] = rnk[q[i]];
  }
  vector<pair<int, int> > ivs;
  for (int i = 0; i < n - 1; i++) {
    if (corr[i] > corr[i + 1]) ivs.push_back({corr[i + 1], corr[i]});
  }
  sort(ivs.begin(), ivs.end());
  int cend = -1, ci = 0;
  char c = 'a';
  int i = 0;
  for (; i < n; i++) {
    if (c == 'a' + k - 1) break;
    if (ci < ivs.size() && ivs[ci].first <= i) {
      cend = max(cend, ivs[ci].second);
      ci++;
    }
    out[p[i]] = c;
    if (cend <= i && i < n - 1) c++;
  }
  for (; i < n; i++) out[p[i]] = c;
  if (c < 'a' + k - 1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int j = 0; j < n; j++) cout << out[j];
    cout << "\n";
  }
  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;
const int MAXN = 200005;
const int MOD = 1000000007;
char out[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(n), q(n), rnk(n), corr(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    rnk[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    q[i]--;
    corr[i] = rnk[q[i]];
  }
  vector<pair<int, int> > ivs;
  for (int i = 0; i < n - 1; i++) {
    if (corr[i] > corr[i + 1]) ivs.push_back({corr[i + 1], corr[i]});
  }
  sort(ivs.begin(), ivs.end());
  int cend = -1, ci = 0;
  char c = 'a';
  int i = 0;
  for (; i < n; i++) {
    if (c == 'a' + k - 1) break;
    if (ci < ivs.size() && ivs[ci].first <= i) {
      cend = max(cend, ivs[ci].second);
      ci++;
    }
    out[p[i]] = c;
    if (cend <= i && i < n - 1) c++;
  }
  for (; i < n; i++) out[p[i]] = c;
  if (c < 'a' + k - 1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int j = 0; j < n; j++) cout << out[j];
    cout << "\n";
  }
  return 0;
}

```