#include <bits/stdc++.h>
const double pi = acos(-1.0);
using namespace std;
inline long long MAX2(long long a, long long b) {
  return (a) > (b) ? (a) : (b);
}
inline long long MAX3(long long a, long long b, long long c) {
  return (a) > (b) ? ((a) > (c) ? (a) : (c)) : ((b) > (c) ? (b) : (c));
}
inline long long MIN2(long long a, long long b) {
  return (a) < (b) ? (a) : (b);
}
inline long long MIN3(long long a, long long b, long long c) {
  return (a) < (b) ? ((a) < (c) ? (a) : (c)) : ((b) < (c) ? (b) : (c));
}
int mod = 1e9 + 7;
int64_t extGcd(int64_t a, int64_t b, int64_t& x, int64_t& y) {
  if (!a) {
    x = 0;
    y = 1;
    return b;
  }
  int64_t x1, y1;
  int64_t d = extGcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
inline long long addmod(long long a, long long b) {
  a = a % mod + b % mod;
  if (a > mod) a %= mod;
  return a;
}
inline long long submod(long long a, long long b) {
  a = a % mod - b % mod;
  if (a < 0) a += mod;
  return a;
}
inline long long mulmod(long long a, long long b) {
  return (a % mod * b % mod) % mod;
}
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
inline long long exp(long long a, long long b) {
  if (a == 0) return 0ll;
  long long r = 1LL;
  while (b > 0) {
    if (b & 1) {
      r = r * (a % mod);
      r = (r + mod) % mod;
    }
    b /= 2;
    a = (a % mod) * (a % mod);
    a = (a + mod) % mod;
  }
  return (r + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  if (a == 0) return b;
  return gcd(b, a % b);
}
unsigned long int setbits(long long n) {
  unsigned long int count = 0;
  while (n) {
    n &= (n - 1);
    count++;
  }
  return count;
}
int main() {
  if (fopen("input.txt", "r"))
    freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  vector<int> g1(n);
  vector<int> g2(n);
  set<int> seen;
  for (int i = 0; i < n; i++) cin >> g1[i];
  for (int i = 0; i < n; i++) cin >> g2[i];
  int p1 = 0;
  int p2 = 0;
  vector<long long> res;
  set<int> s2;
  set<int> s1;
  for (int i = 0; i < n; i++) {
    g1[i]--;
    g2[i]--;
    if (s2.count(g1[i])) {
      s2.erase(g1[i]);
    } else {
      s1.insert(g1[i]);
    }
    if (s1.count(g2[i])) {
      s1.erase(g2[i]);
    } else {
      s2.insert(g2[i]);
    }
    if (s1.empty() and s2.empty()) {
      res.push_back(i);
    }
  }
  vector<char> ans(n);
  if (res.size() < k) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  char x = 'a';
  int prev = 0;
  for (int i = 0; i < (int)res.size(); i++) {
    int curr = res[i];
    int start = prev;
    int end = curr;
    prev = curr + 1;
    for (int j = start; j <= end; j++) {
      ans[g1[j]] = x;
    }
    if (x != 'z') x++;
  }
  for (auto x : ans) cout << x;
  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>
const double pi = acos(-1.0);
using namespace std;
inline long long MAX2(long long a, long long b) {
  return (a) > (b) ? (a) : (b);
}
inline long long MAX3(long long a, long long b, long long c) {
  return (a) > (b) ? ((a) > (c) ? (a) : (c)) : ((b) > (c) ? (b) : (c));
}
inline long long MIN2(long long a, long long b) {
  return (a) < (b) ? (a) : (b);
}
inline long long MIN3(long long a, long long b, long long c) {
  return (a) < (b) ? ((a) < (c) ? (a) : (c)) : ((b) < (c) ? (b) : (c));
}
int mod = 1e9 + 7;
int64_t extGcd(int64_t a, int64_t b, int64_t& x, int64_t& y) {
  if (!a) {
    x = 0;
    y = 1;
    return b;
  }
  int64_t x1, y1;
  int64_t d = extGcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
inline long long addmod(long long a, long long b) {
  a = a % mod + b % mod;
  if (a > mod) a %= mod;
  return a;
}
inline long long submod(long long a, long long b) {
  a = a % mod - b % mod;
  if (a < 0) a += mod;
  return a;
}
inline long long mulmod(long long a, long long b) {
  return (a % mod * b % mod) % mod;
}
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
inline long long exp(long long a, long long b) {
  if (a == 0) return 0ll;
  long long r = 1LL;
  while (b > 0) {
    if (b & 1) {
      r = r * (a % mod);
      r = (r + mod) % mod;
    }
    b /= 2;
    a = (a % mod) * (a % mod);
    a = (a + mod) % mod;
  }
  return (r + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  if (a == 0) return b;
  return gcd(b, a % b);
}
unsigned long int setbits(long long n) {
  unsigned long int count = 0;
  while (n) {
    n &= (n - 1);
    count++;
  }
  return count;
}
int main() {
  if (fopen("input.txt", "r"))
    freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  vector<int> g1(n);
  vector<int> g2(n);
  set<int> seen;
  for (int i = 0; i < n; i++) cin >> g1[i];
  for (int i = 0; i < n; i++) cin >> g2[i];
  int p1 = 0;
  int p2 = 0;
  vector<long long> res;
  set<int> s2;
  set<int> s1;
  for (int i = 0; i < n; i++) {
    g1[i]--;
    g2[i]--;
    if (s2.count(g1[i])) {
      s2.erase(g1[i]);
    } else {
      s1.insert(g1[i]);
    }
    if (s1.count(g2[i])) {
      s1.erase(g2[i]);
    } else {
      s2.insert(g2[i]);
    }
    if (s1.empty() and s2.empty()) {
      res.push_back(i);
    }
  }
  vector<char> ans(n);
  if (res.size() < k) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  char x = 'a';
  int prev = 0;
  for (int i = 0; i < (int)res.size(); i++) {
    int curr = res[i];
    int start = prev;
    int end = curr;
    prev = curr + 1;
    for (int j = start; j <= end; j++) {
      ans[g1[j]] = x;
    }
    if (x != 'z') x++;
  }
  for (auto x : ans) cout << x;
  return 0;
}

```