#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9 + 7;
void fast() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
}
struct Node {
  int sum;
};
vector<Node> mas;
void build(int v, int l, int r) {
  if (l + 1 == r) {
    mas[v].sum = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  mas[v].sum = 0;
  return;
}
void update(int v, int l, int r, int ql, int qr) {
  if (l >= qr || r <= ql) {
    return;
  }
  if (l + 1 == r) {
    mas[v].sum = 1 - mas[v].sum;
    return;
  }
  int m = (l + r) / 2;
  update(2 * v + 1, l, m, ql, qr);
  update(2 * v + 2, m, r, ql, qr);
  mas[v].sum = mas[2 * v + 1].sum | mas[2 * v + 2].sum;
  return;
}
int query(int v, int l, int r, int ql, int qr) { return mas[v].sum; }
signed main() {
  fast();
  int n, k;
  cin >> n >> k;
  vector<int> m1(n);
  vector<int> m2(n);
  vector<int> val;
  vector<int> res(n + 1);
  mas.resize(4 * n);
  build(0, 0, n + 1);
  for (int i = 0; i < n; i++) {
    cin >> m1[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> m2[i];
  }
  for (int i = 0; i < n; i++) {
    int x = m1[i];
    int y = m2[i];
    update(0, 0, n + 1, x, x + 1);
    update(0, 0, n + 1, y, y + 1);
    if (query(0, 0, n + 1, 0, n + 1) == 0) {
      val.push_back(i);
    }
  }
  if (val.size() < k) {
    cout << "NO";
    return 0;
  }
  int cnt = 0;
  int l = 0;
  for (auto el : val) {
    while (l <= el) {
      res[m1[l]] = cnt;
      l++;
    }
    cnt++;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    if (res[i] > 25) {
      res[i] = 25;
    }
    cout << char(res[i] + 'a');
  }
  return 0;
}

### 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;
const int INF = 2e9 + 7;
void fast() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
}
struct Node {
  int sum;
};
vector<Node> mas;
void build(int v, int l, int r) {
  if (l + 1 == r) {
    mas[v].sum = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  mas[v].sum = 0;
  return;
}
void update(int v, int l, int r, int ql, int qr) {
  if (l >= qr || r <= ql) {
    return;
  }
  if (l + 1 == r) {
    mas[v].sum = 1 - mas[v].sum;
    return;
  }
  int m = (l + r) / 2;
  update(2 * v + 1, l, m, ql, qr);
  update(2 * v + 2, m, r, ql, qr);
  mas[v].sum = mas[2 * v + 1].sum | mas[2 * v + 2].sum;
  return;
}
int query(int v, int l, int r, int ql, int qr) { return mas[v].sum; }
signed main() {
  fast();
  int n, k;
  cin >> n >> k;
  vector<int> m1(n);
  vector<int> m2(n);
  vector<int> val;
  vector<int> res(n + 1);
  mas.resize(4 * n);
  build(0, 0, n + 1);
  for (int i = 0; i < n; i++) {
    cin >> m1[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> m2[i];
  }
  for (int i = 0; i < n; i++) {
    int x = m1[i];
    int y = m2[i];
    update(0, 0, n + 1, x, x + 1);
    update(0, 0, n + 1, y, y + 1);
    if (query(0, 0, n + 1, 0, n + 1) == 0) {
      val.push_back(i);
    }
  }
  if (val.size() < k) {
    cout << "NO";
    return 0;
  }
  int cnt = 0;
  int l = 0;
  for (auto el : val) {
    while (l <= el) {
      res[m1[l]] = cnt;
      l++;
    }
    cnt++;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    if (res[i] > 25) {
      res[i] = 25;
    }
    cout << char(res[i] + 'a');
  }
  return 0;
}

```