#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powmod(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = powmod(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modif(long long x, long long m) { return (powmod(x, m - 2, m)); }
bool fa(vector<long long> &x, vector<long long> &y) { return x[0] < y[0]; }
bool fa1(vector<long long> &x, vector<long long> &y) { return x[1] < y[1]; }
bool f1(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.second < y.second;
}
bool f2(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.first < y.first;
}
bool f3(long long &x, long long &y) { return x > y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long b[n], c[n];
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
  }
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    x--;
    c[x] = i;
  }
  long long l = -1;
  vector<long long> h;
  vector<vector<long long> > fin;
  vector<long long> ro;
  fin.push_back(ro);
  for (int i = 0; i < n; i++) {
    l = max(l, c[b[i]]);
    fin[fin.size() - 1].push_back(b[i]);
    if (i >= l) {
      h.push_back(i);
      fin.push_back(ro);
    }
  }
  if (h.size() < k) {
    cout << "NO";
    exit(0);
  }
  long long t = 0;
  char meow = 'a';
  string ans(n, 'a');
  for (int i = 0; i < fin.size(); i++) {
    for (auto x : fin[i]) {
      ans[x] = meow;
    }
    meow++;
    meow = min(meow, 'z');
  }
  cout << "YES\n";
  cout << ans;
  return 0;
}

### 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;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powmod(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = powmod(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modif(long long x, long long m) { return (powmod(x, m - 2, m)); }
bool fa(vector<long long> &x, vector<long long> &y) { return x[0] < y[0]; }
bool fa1(vector<long long> &x, vector<long long> &y) { return x[1] < y[1]; }
bool f1(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.second < y.second;
}
bool f2(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.first < y.first;
}
bool f3(long long &x, long long &y) { return x > y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long b[n], c[n];
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
  }
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    x--;
    c[x] = i;
  }
  long long l = -1;
  vector<long long> h;
  vector<vector<long long> > fin;
  vector<long long> ro;
  fin.push_back(ro);
  for (int i = 0; i < n; i++) {
    l = max(l, c[b[i]]);
    fin[fin.size() - 1].push_back(b[i]);
    if (i >= l) {
      h.push_back(i);
      fin.push_back(ro);
    }
  }
  if (h.size() < k) {
    cout << "NO";
    exit(0);
  }
  long long t = 0;
  char meow = 'a';
  string ans(n, 'a');
  for (int i = 0; i < fin.size(); i++) {
    for (auto x : fin[i]) {
      ans[x] = meow;
    }
    meow++;
    meow = min(meow, 'z');
  }
  cout << "YES\n";
  cout << ans;
  return 0;
}

```