#include <bits/stdc++.h>
using namespace std;
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(long long &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const long long &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 1e6 + 10;
int a[SIZE], b[SIZE], d[SIZE];
char AA;
char an[SIZE];
int main() {
  int n, k;
  R(n, k);
  for (int i = (1); i <= (n); ++i) R(a[i]);
  for (int i = (1); i <= (n); ++i) R(b[i]);
  AA = 'a';
  for (int i = 1, j; i <= n; i = j) {
    k--;
    int v = 0;
    for (j = i;; j++) {
      if (d[a[j]] == 0) {
        v++;
        d[a[j]]++;
      } else if (d[a[j]] == 1) {
        v--;
        d[a[j]]--;
      }
      if (d[b[j]] == 0) {
        v++;
        d[b[j]]++;
      } else if (d[b[j]] == 1) {
        v--;
        d[b[j]]--;
      }
      if (!v) {
        j++;
        break;
      }
    }
    for (int r = (i); r < (j); ++r) {
      an[a[r]] = AA;
    }
    if (AA < 'z') AA++;
  }
  if (k <= 0) {
    W("YES");
    W(an + 1);
  } else {
    W("NO");
  }
  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;
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(long long &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const long long &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 1e6 + 10;
int a[SIZE], b[SIZE], d[SIZE];
char AA;
char an[SIZE];
int main() {
  int n, k;
  R(n, k);
  for (int i = (1); i <= (n); ++i) R(a[i]);
  for (int i = (1); i <= (n); ++i) R(b[i]);
  AA = 'a';
  for (int i = 1, j; i <= n; i = j) {
    k--;
    int v = 0;
    for (j = i;; j++) {
      if (d[a[j]] == 0) {
        v++;
        d[a[j]]++;
      } else if (d[a[j]] == 1) {
        v--;
        d[a[j]]--;
      }
      if (d[b[j]] == 0) {
        v++;
        d[b[j]]++;
      } else if (d[b[j]] == 1) {
        v--;
        d[b[j]]--;
      }
      if (!v) {
        j++;
        break;
      }
    }
    for (int r = (i); r < (j); ++r) {
      an[a[r]] = AA;
    }
    if (AA < 'z') AA++;
  }
  if (k <= 0) {
    W("YES");
    W(an + 1);
  } else {
    W("NO");
  }
  return 0;
}

```