#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int a[maxn], b[maxn];
char c[maxn] = {0};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]--;
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), b[i]--;
  set<int> st1, st2;
  vector<int> vt;
  for (int i = 1; i <= n; i++) {
    if (st1.count(b[i]))
      st1.erase(b[i]);
    else
      st2.insert(b[i]);
    if (st2.count(a[i]))
      st2.erase(a[i]);
    else
      st1.insert(a[i]);
    if (st1.empty() && st2.empty()) vt.push_back(i);
  }
  if (vt.size() < k) {
    printf("NO\n");
    return 0;
  }
  int l = 0;
  char s = 'a';
  for (int i = 0; i < vt.size(); i++) {
    for (int j = l; j <= vt[i]; j++) c[a[j]] = s;
    l = vt[i] + 1;
    s += 1;
    s = min('z', s);
  }
  printf("YES\n");
  printf("%s\n", c);
}

### 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 maxn = 2e5 + 10;
int a[maxn], b[maxn];
char c[maxn] = {0};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]--;
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), b[i]--;
  set<int> st1, st2;
  vector<int> vt;
  for (int i = 1; i <= n; i++) {
    if (st1.count(b[i]))
      st1.erase(b[i]);
    else
      st2.insert(b[i]);
    if (st2.count(a[i]))
      st2.erase(a[i]);
    else
      st1.insert(a[i]);
    if (st1.empty() && st2.empty()) vt.push_back(i);
  }
  if (vt.size() < k) {
    printf("NO\n");
    return 0;
  }
  int l = 0;
  char s = 'a';
  for (int i = 0; i < vt.size(); i++) {
    for (int j = l; j <= vt[i]; j++) c[a[j]] = s;
    l = vt[i] + 1;
    s += 1;
    s = min('z', s);
  }
  printf("YES\n");
  printf("%s\n", c);
}

```