#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k, a[200010], b[200010], fa[200010], pos[200010], cnt = 0;
char ch[200010];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  int top1 = 1, top2 = 1;
  while (1) {
    if (top1 > n || top2 > n) break;
    int fx = find(a[top1]), fy = find(b[top2]);
    if (fx == fy) {
      if (top1 == top2) {
        pos[++cnt] = top1;
        top1++, top2++;
      } else
        top2++;
    } else
      fa[find(a[top1])] = find(a[top1 + 1]), top1++;
  }
  if (cnt < k) return printf("NO\n"), 0;
  int last = 1, tot = 1, tmp = 0;
  while (1) {
    if (tot > cnt) break;
    for (int i = last; i <= pos[tot]; i++) {
      ch[a[i]] = tmp + 'a';
    }
    last = pos[tot] + 1;
    tot++;
    if (tot <= k) tmp++;
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%c", ch[i]);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k, a[200010], b[200010], fa[200010], pos[200010], cnt = 0;
char ch[200010];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  int top1 = 1, top2 = 1;
  while (1) {
    if (top1 > n || top2 > n) break;
    int fx = find(a[top1]), fy = find(b[top2]);
    if (fx == fy) {
      if (top1 == top2) {
        pos[++cnt] = top1;
        top1++, top2++;
      } else
        top2++;
    } else
      fa[find(a[top1])] = find(a[top1 + 1]), top1++;
  }
  if (cnt < k) return printf("NO\n"), 0;
  int last = 1, tot = 1, tmp = 0;
  while (1) {
    if (tot > cnt) break;
    for (int i = last; i <= pos[tot]; i++) {
      ch[a[i]] = tmp + 'a';
    }
    last = pos[tot] + 1;
    tot++;
    if (tot <= k) tmp++;
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%c", ch[i]);
  }
  return 0;
}

```