#include <bits/stdc++.h>
using namespace std;
int p[int(1e5 * 2 + 99)], q[int(1e5 * 2 + 99)];
vector<int> g[int(1e5 * 2 + 99)], rev[int(1e5 * 2 + 99)];
bool was[int(1e5 * 2 + 99)];
vector<int> order;
void dfs(int v) {
  was[v] = true;
  for (int i : g[v]) {
    if (!was[i]) {
      dfs(i);
    }
  }
  order.push_back(v);
}
vector<char> main_str;
void dfs_rev(int v, char sym) {
  was[v] = true;
  for (int i : rev[v]) {
    if (!was[i]) {
      dfs_rev(i, sym);
    }
  }
  main_str[v] = sym;
}
int main() {
  int n, k;
  scanf("%i%i", &n, &k);
  main_str.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%i", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    scanf("%i", &q[i]);
    q[i]--;
  }
  for (int i = 1; i < n; i++) {
    g[p[i]].push_back(p[i - 1]);
    rev[p[i - 1]].push_back(p[i]);
  }
  for (int i = 1; i < n; i++) {
    g[q[i]].push_back(q[i - 1]);
    rev[q[i - 1]].push_back(q[i]);
  }
  for (int i = 0; i < n; i++) {
    if (!was[i]) {
      dfs(i);
    }
  }
  fill(was, was + n, 0);
  reverse(order.begin(), order.end());
  char main_char = 'a' + k - 1;
  int count_ = 0;
  for (int i = 0; i < n; i++) {
    if (!was[order[i]]) {
      dfs_rev(order[i], main_char);
      count_++;
      main_char--;
      if (main_char <= 'a') main_char = 'a';
    }
  }
  if (count_ < k) {
    printf("NO");
  } else {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      printf("%c", main_str[i]);
    }
  }
}

### Prompt

Your challenge is to write 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 p[int(1e5 * 2 + 99)], q[int(1e5 * 2 + 99)];
vector<int> g[int(1e5 * 2 + 99)], rev[int(1e5 * 2 + 99)];
bool was[int(1e5 * 2 + 99)];
vector<int> order;
void dfs(int v) {
  was[v] = true;
  for (int i : g[v]) {
    if (!was[i]) {
      dfs(i);
    }
  }
  order.push_back(v);
}
vector<char> main_str;
void dfs_rev(int v, char sym) {
  was[v] = true;
  for (int i : rev[v]) {
    if (!was[i]) {
      dfs_rev(i, sym);
    }
  }
  main_str[v] = sym;
}
int main() {
  int n, k;
  scanf("%i%i", &n, &k);
  main_str.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%i", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    scanf("%i", &q[i]);
    q[i]--;
  }
  for (int i = 1; i < n; i++) {
    g[p[i]].push_back(p[i - 1]);
    rev[p[i - 1]].push_back(p[i]);
  }
  for (int i = 1; i < n; i++) {
    g[q[i]].push_back(q[i - 1]);
    rev[q[i - 1]].push_back(q[i]);
  }
  for (int i = 0; i < n; i++) {
    if (!was[i]) {
      dfs(i);
    }
  }
  fill(was, was + n, 0);
  reverse(order.begin(), order.end());
  char main_char = 'a' + k - 1;
  int count_ = 0;
  for (int i = 0; i < n; i++) {
    if (!was[order[i]]) {
      dfs_rev(order[i], main_char);
      count_++;
      main_char--;
      if (main_char <= 'a') main_char = 'a';
    }
  }
  if (count_ < k) {
    printf("NO");
  } else {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      printf("%c", main_str[i]);
    }
  }
}

```