#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> mp;
map<long long int, long long int> mp1;
long long int p[200005];
long long int q[200005];
char ans[200005];
vector<long long int> vec;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, k, i, f = 0, prep = 1, preq = 1, pos, val, j = 1, max1, max2;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> p[i];
    mp1[p[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    cin >> q[i];
    mp[q[i]] = i;
  }
  i = 1;
  j = 1;
  while (i <= n || j <= n) {
    max1 = preq;
    max2 = prep;
    while (i <= prep) {
      pos = mp[p[i]];
      max1 = max(max1, pos);
      i++;
    }
    if (max1 == preq) {
      vec.push_back(preq);
      f++;
      prep++;
      j = preq + 1;
    } else {
      while (j <= max1) {
        pos = mp1[q[j]];
        max2 = max(pos, max2);
        j++;
      }
      preq = max1;
      if (max2 == prep) {
        vec.push_back(preq);
        prep++;
        f++;
      } else
        prep = max2;
    }
  }
  if (f >= k) {
    cout << "YES" << endl;
    i = 1;
    pos = 0;
    val = 97;
    while (i <= n) {
      if (pos < vec.size() && i > vec[pos]) {
        pos++;
        val = min(val + 1, 122ll);
      }
      ans[q[i]] = (char)(val);
      i++;
    }
    for (i = 1; i <= n; i++) cout << ans[i];
  } else
    cout << "NO" << endl;
}

### 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;
map<long long int, long long int> mp;
map<long long int, long long int> mp1;
long long int p[200005];
long long int q[200005];
char ans[200005];
vector<long long int> vec;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, k, i, f = 0, prep = 1, preq = 1, pos, val, j = 1, max1, max2;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> p[i];
    mp1[p[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    cin >> q[i];
    mp[q[i]] = i;
  }
  i = 1;
  j = 1;
  while (i <= n || j <= n) {
    max1 = preq;
    max2 = prep;
    while (i <= prep) {
      pos = mp[p[i]];
      max1 = max(max1, pos);
      i++;
    }
    if (max1 == preq) {
      vec.push_back(preq);
      f++;
      prep++;
      j = preq + 1;
    } else {
      while (j <= max1) {
        pos = mp1[q[j]];
        max2 = max(pos, max2);
        j++;
      }
      preq = max1;
      if (max2 == prep) {
        vec.push_back(preq);
        prep++;
        f++;
      } else
        prep = max2;
    }
  }
  if (f >= k) {
    cout << "YES" << endl;
    i = 1;
    pos = 0;
    val = 97;
    while (i <= n) {
      if (pos < vec.size() && i > vec[pos]) {
        pos++;
        val = min(val + 1, 122ll);
      }
      ans[q[i]] = (char)(val);
      i++;
    }
    for (i = 1; i <= n; i++) cout << ans[i];
  } else
    cout << "NO" << endl;
}

```