#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void MIN(T &a, T b) {
  if (a > b) a = b;
}
template <typename T>
inline void MAX(T &a, T b) {
  if (a < b) a = b;
}
const int N = 200005;
int n, k, dd[N], p[N];
char ans[N];
pair<int, int> a[N * 2], b[N * 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) {
    cin >> p[i];
    dd[p[i]] = i;
  }
  int prv, now, cnt = 0;
  for (int i = (1); i <= (n); ++i) {
    if (i >= 2) prv = now;
    cin >> now;
    a[++cnt] = make_pair(now, now);
    if ((i >= 2) && (dd[prv] > dd[now])) a[++cnt] = make_pair(dd[now], dd[prv]);
  }
  sort(a + 1, a + cnt + 1, [](pair<int, int> x, pair<int, int> y) {
    return (x.first < y.first) ||
           ((x.first == y.first) && (x.second > y.second));
  });
  int newCnt = 0, id = 1;
  while (id <= cnt) {
    int pv = id;
    for (; id + 1 <= cnt && a[id + 1].second <= a[pv].second; ++id)
      ;
    b[++newCnt] = a[pv];
    ++id;
  }
  int Union = 0;
  id = 1;
  while (id <= newCnt) {
    int pv = id;
    for (; id + 1 <= newCnt && b[id + 1].first <= b[id].second; ++id)
      ;
    b[++Union] = make_pair(b[pv].first, b[id].second);
    ++id;
  }
  if (Union < k) return cout << "NO", 0;
  cout << "YES\n";
  char x = 'a' - 1;
  for (int i = (1); i <= (Union); ++i) {
    x = min(char((int)x + 1), 'z');
    for (int l = (b[i].first); l <= (b[i].second); ++l) ans[p[l]] = char(x);
  }
  for (int i = (1); i <= (n); ++i) cout << ans[i];
  return 0;
}

### Prompt

Please create a solution in CPP 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 <typename T>
inline void MIN(T &a, T b) {
  if (a > b) a = b;
}
template <typename T>
inline void MAX(T &a, T b) {
  if (a < b) a = b;
}
const int N = 200005;
int n, k, dd[N], p[N];
char ans[N];
pair<int, int> a[N * 2], b[N * 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) {
    cin >> p[i];
    dd[p[i]] = i;
  }
  int prv, now, cnt = 0;
  for (int i = (1); i <= (n); ++i) {
    if (i >= 2) prv = now;
    cin >> now;
    a[++cnt] = make_pair(now, now);
    if ((i >= 2) && (dd[prv] > dd[now])) a[++cnt] = make_pair(dd[now], dd[prv]);
  }
  sort(a + 1, a + cnt + 1, [](pair<int, int> x, pair<int, int> y) {
    return (x.first < y.first) ||
           ((x.first == y.first) && (x.second > y.second));
  });
  int newCnt = 0, id = 1;
  while (id <= cnt) {
    int pv = id;
    for (; id + 1 <= cnt && a[id + 1].second <= a[pv].second; ++id)
      ;
    b[++newCnt] = a[pv];
    ++id;
  }
  int Union = 0;
  id = 1;
  while (id <= newCnt) {
    int pv = id;
    for (; id + 1 <= newCnt && b[id + 1].first <= b[id].second; ++id)
      ;
    b[++Union] = make_pair(b[pv].first, b[id].second);
    ++id;
  }
  if (Union < k) return cout << "NO", 0;
  cout << "YES\n";
  char x = 'a' - 1;
  for (int i = (1); i <= (Union); ++i) {
    x = min(char((int)x + 1), 'z');
    for (int l = (b[i].first); l <= (b[i].second); ++l) ans[p[l]] = char(x);
  }
  for (int i = (1); i <= (n); ++i) cout << ans[i];
  return 0;
}

```