#include <bits/stdc++.h>
using namespace std;
int n, k, ks, cnt, lena, lenb;
int a[200005], b[200005], ans[200005];
char c[200005];
bool visa[200005], visb[200005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= n; i++) {
    if (visb[a[i]] == 1) {
      visb[a[i]] = 0;
      lenb--;
    } else {
      visa[a[i]] = 1;
      lena++;
    }
    if (visa[b[i]] == 1) {
      visa[b[i]] = 0;
      lena--;
    } else {
      visb[b[i]] = 1;
      lenb++;
    }
    if (lena == 0 && lenb == 0) {
      cnt++;
      ans[cnt] = i;
    }
  }
  if (cnt < k) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  ks = 1;
  int r;
  char ch;
  for (int i = 1; i <= cnt; i++) {
    r = ans[i];
    ch = 'a' + min(i, 26) - 1;
    for (int j = ks; j <= r; j++) c[a[j]] = ch;
    ks = r + 1;
  }
  printf("%s", c + 1);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int n, k, ks, cnt, lena, lenb;
int a[200005], b[200005], ans[200005];
char c[200005];
bool visa[200005], visb[200005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= n; i++) {
    if (visb[a[i]] == 1) {
      visb[a[i]] = 0;
      lenb--;
    } else {
      visa[a[i]] = 1;
      lena++;
    }
    if (visa[b[i]] == 1) {
      visa[b[i]] = 0;
      lena--;
    } else {
      visb[b[i]] = 1;
      lenb++;
    }
    if (lena == 0 && lenb == 0) {
      cnt++;
      ans[cnt] = i;
    }
  }
  if (cnt < k) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  ks = 1;
  int r;
  char ch;
  for (int i = 1; i <= cnt; i++) {
    r = ans[i];
    ch = 'a' + min(i, 26) - 1;
    for (int j = ks; j <= r; j++) c[a[j]] = ch;
    ks = r + 1;
  }
  printf("%s", c + 1);
  return 0;
}

```