#include <bits/stdc++.h>
using namespace std;
int n, k;
int p[200010], q[200010];
int compo[200010];
vector<int> gr;
void dfs(int pos, int v) {
  compo[pos] = v;
  if (compo[gr[pos]] == 0) dfs(gr[pos], v);
}
int main() {
  scanf("%d %d", &n, &k);
  gr.resize(n + 1);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < n; i++) {
    scanf("%d", &q[i]);
    gr[p[i]] = q[i];
  }
  int tam = 1;
  for (int i = 1; i <= n; i++) {
    if (compo[i] == 0) {
      dfs(i, tam);
      tam++;
    }
  }
  vector<pair<int, int> > corto;
  for (int i = 0; i < n; i++) corto.push_back({compo[p[i]], i});
  sort(corto.begin(), corto.end());
  vector<pair<int, int> > rangos;
  rangos.push_back({corto[0].second, corto[0].second});
  for (int i = 1; i < n; i++) {
    if (corto[i - 1].first != corto[i].first)
      rangos.push_back({corto[i].second, corto[i].second});
    else
      rangos.back().second = corto[i].second;
  }
  int resp[n];
  sort(rangos.begin(), rangos.end());
  int fin = -1;
  int val = 0;
  for (int i = 0; i < (int)rangos.size(); i++) {
    if (rangos[i].first > fin) val++;
    while (fin < rangos[i].second) {
      fin++;
      resp[fin] = val;
    }
  }
  char salida[n + 10];
  if (k <= val) {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      int aux = resp[i] - 1 + (int)'a';
      if (aux <= (int)'z')
        salida[p[i] - 1] = (char)aux;
      else
        salida[p[i] - 1] = 'z';
    }
    salida[n] = '\0';
    printf("%s\n", salida);
  } else
    printf("NO\n");
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int n, k;
int p[200010], q[200010];
int compo[200010];
vector<int> gr;
void dfs(int pos, int v) {
  compo[pos] = v;
  if (compo[gr[pos]] == 0) dfs(gr[pos], v);
}
int main() {
  scanf("%d %d", &n, &k);
  gr.resize(n + 1);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < n; i++) {
    scanf("%d", &q[i]);
    gr[p[i]] = q[i];
  }
  int tam = 1;
  for (int i = 1; i <= n; i++) {
    if (compo[i] == 0) {
      dfs(i, tam);
      tam++;
    }
  }
  vector<pair<int, int> > corto;
  for (int i = 0; i < n; i++) corto.push_back({compo[p[i]], i});
  sort(corto.begin(), corto.end());
  vector<pair<int, int> > rangos;
  rangos.push_back({corto[0].second, corto[0].second});
  for (int i = 1; i < n; i++) {
    if (corto[i - 1].first != corto[i].first)
      rangos.push_back({corto[i].second, corto[i].second});
    else
      rangos.back().second = corto[i].second;
  }
  int resp[n];
  sort(rangos.begin(), rangos.end());
  int fin = -1;
  int val = 0;
  for (int i = 0; i < (int)rangos.size(); i++) {
    if (rangos[i].first > fin) val++;
    while (fin < rangos[i].second) {
      fin++;
      resp[fin] = val;
    }
  }
  char salida[n + 10];
  if (k <= val) {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      int aux = resp[i] - 1 + (int)'a';
      if (aux <= (int)'z')
        salida[p[i] - 1] = (char)aux;
      else
        salida[p[i] - 1] = 'z';
    }
    salida[n] = '\0';
    printf("%s\n", salida);
  } else
    printf("NO\n");
  return 0;
}

```