#include <bits/stdc++.h>
using namespace std;
const int INF = (1 << 30) - 1;
const long double PI = 3.1415926535898;
const long long oo = (1ll << 62) - 1;
const int N = 2e5 + 5;
int n, k;
int p[N], q[N], l[N], ans[N], res[N];
vector<pair<int, int> > v, cnt;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) cin >> q[i];
  for (int i = 1; i <= n; i++) {
    l[p[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    int a = l[q[i]], b = l[q[i + 1]];
    if (a > b) v.push_back(make_pair(b, a));
  }
  sort(v.begin(), v.end());
  cnt.push_back(v.front());
  for (int i = 0; i < v.size(); i++) {
    if (v[i].first <= cnt.back().second) {
      cnt.back().second = max(cnt.back().second, v[i].second);
    } else {
      cnt.push_back(v[i]);
    }
  }
  int j = 0;
  ans[1] = 1;
  for (int i = 2; i <= n; i++) {
    while (i > cnt[j].second && j < cnt.size()) j++;
    if (i > cnt[j].first && i <= cnt[j].second) {
      ans[i] = ans[i - 1];
    } else
      ans[i] = ans[i - 1] + 1;
  }
  for (int i = 1; i <= n; i++) {
    res[p[i]] = min(k, ans[i]);
  }
  if (ans[n] < k) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) {
    cout << char(res[i] - 1 + 'a');
  }
}

### 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;
const int INF = (1 << 30) - 1;
const long double PI = 3.1415926535898;
const long long oo = (1ll << 62) - 1;
const int N = 2e5 + 5;
int n, k;
int p[N], q[N], l[N], ans[N], res[N];
vector<pair<int, int> > v, cnt;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) cin >> q[i];
  for (int i = 1; i <= n; i++) {
    l[p[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    int a = l[q[i]], b = l[q[i + 1]];
    if (a > b) v.push_back(make_pair(b, a));
  }
  sort(v.begin(), v.end());
  cnt.push_back(v.front());
  for (int i = 0; i < v.size(); i++) {
    if (v[i].first <= cnt.back().second) {
      cnt.back().second = max(cnt.back().second, v[i].second);
    } else {
      cnt.push_back(v[i]);
    }
  }
  int j = 0;
  ans[1] = 1;
  for (int i = 2; i <= n; i++) {
    while (i > cnt[j].second && j < cnt.size()) j++;
    if (i > cnt[j].first && i <= cnt[j].second) {
      ans[i] = ans[i - 1];
    } else
      ans[i] = ans[i - 1] + 1;
  }
  for (int i = 1; i <= n; i++) {
    res[p[i]] = min(k, ans[i]);
  }
  if (ans[n] < k) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) {
    cout << char(res[i] - 1 + 'a');
  }
}

```