#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
const int INF = 0x3f3f3f3f;
const int iinf = 1 << 30;
const long long linf = 2e18;
const int mod = 1000000007;
const double eps = 1e-7;
void douout(double x) { printf("%lf\n", x + 0.0000000001); }
template <class T>
void print(T a) {
  cout << a << endl;
  exit(0);
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void Tmin(T &a, T &b, T &c) {
  return min(a, min(b, c));
}
template <class T>
void Tmax(T &a, T &b, T &c) {
  return max(a, max(b, c));
}
void add(int &a, int b) { a = a + b < mod ? a + b : a + b - mod; }
void sub(int &a, int b) { a = (a - b + mod) % mod; }
void mul(int &a, int b) { a = (long long)a * b % mod; }
int addv(int a, int b) { return (a += b) >= mod ? a -= mod : a; }
int subv(int a, int b) { return (a -= b) < 0 ? a += mod : a; }
int mulv(int a, int b) { return (long long)a * b % mod; }
int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int pw(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) s = (long long)s * a % mod;
  return s;
}
int n, k, cnt, l = 1;
int a[N], b[N];
bool t[N];
char c[N];
signed main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  for (int i = (1); i <= (n); i++) scanf("%d", &b[i]);
  for (int i = (1); i <= (n); i++) {
    c[a[i]] = c[b[i]] = 'a' + l - 1;
    if (t[a[i]])
      cnt--;
    else
      cnt++;
    t[a[i]] ^= 1;
    if (t[b[i]])
      cnt--;
    else
      cnt++;
    t[b[i]] ^= 1;
    if (cnt == 0) l++;
    if (l > k) {
      for (int i = (1); i <= (n); i++)
        if (c[i] == 0) c[i] = 'a' + k - 1;
      break;
    }
  }
  if (l <= k)
    printf("NO");
  else
    printf("YES\n%s", c + 1);
}

### 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;
const int N = 200010;
const int INF = 0x3f3f3f3f;
const int iinf = 1 << 30;
const long long linf = 2e18;
const int mod = 1000000007;
const double eps = 1e-7;
void douout(double x) { printf("%lf\n", x + 0.0000000001); }
template <class T>
void print(T a) {
  cout << a << endl;
  exit(0);
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void Tmin(T &a, T &b, T &c) {
  return min(a, min(b, c));
}
template <class T>
void Tmax(T &a, T &b, T &c) {
  return max(a, max(b, c));
}
void add(int &a, int b) { a = a + b < mod ? a + b : a + b - mod; }
void sub(int &a, int b) { a = (a - b + mod) % mod; }
void mul(int &a, int b) { a = (long long)a * b % mod; }
int addv(int a, int b) { return (a += b) >= mod ? a -= mod : a; }
int subv(int a, int b) { return (a -= b) < 0 ? a += mod : a; }
int mulv(int a, int b) { return (long long)a * b % mod; }
int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int pw(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) s = (long long)s * a % mod;
  return s;
}
int n, k, cnt, l = 1;
int a[N], b[N];
bool t[N];
char c[N];
signed main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  for (int i = (1); i <= (n); i++) scanf("%d", &b[i]);
  for (int i = (1); i <= (n); i++) {
    c[a[i]] = c[b[i]] = 'a' + l - 1;
    if (t[a[i]])
      cnt--;
    else
      cnt++;
    t[a[i]] ^= 1;
    if (t[b[i]])
      cnt--;
    else
      cnt++;
    t[b[i]] ^= 1;
    if (cnt == 0) l++;
    if (l > k) {
      for (int i = (1); i <= (n); i++)
        if (c[i] == 0) c[i] = 'a' + k - 1;
      break;
    }
  }
  if (l <= k)
    printf("NO");
  else
    printf("YES\n%s", c + 1);
}

```