#include <bits/stdc++.h>
using namespace std;
int n, k, a[200010], b[200010], fa[200010], c[200010], s = 0;
int i1 = 1, i2 = 1;
int las = 1, xb = 1, t = 0;
char ch[200010];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  while (true) {
    if (i1 > n || i2 > n) break;
    int fx = find(a[i1]), fy = find(b[i2]);
    if (fx == fy) {
      if (i1 == i2) {
        c[++s] = i1;
        i1++;
        i2++;
      } else
        i2++;
    } else
      fa[find(a[i1])] = find(a[i1 + 1]), i1++;
  }
  if (s < k) {
    printf("NO\n");
    return 0;
  }
  while (true) {
    if (xb > s) {
      break;
    }
    for (int i = las; i <= c[xb]; i++) ch[a[i]] = t + 'a';
    las = c[xb++] + 1;
    if (xb <= k) t++;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << ch[i];
  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;
int n, k, a[200010], b[200010], fa[200010], c[200010], s = 0;
int i1 = 1, i2 = 1;
int las = 1, xb = 1, t = 0;
char ch[200010];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  while (true) {
    if (i1 > n || i2 > n) break;
    int fx = find(a[i1]), fy = find(b[i2]);
    if (fx == fy) {
      if (i1 == i2) {
        c[++s] = i1;
        i1++;
        i2++;
      } else
        i2++;
    } else
      fa[find(a[i1])] = find(a[i1 + 1]), i1++;
  }
  if (s < k) {
    printf("NO\n");
    return 0;
  }
  while (true) {
    if (xb > s) {
      break;
    }
    for (int i = las; i <= c[xb]; i++) ch[a[i]] = t + 'a';
    las = c[xb++] + 1;
    if (xb <= k) t++;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << ch[i];
  return 0;
}

```