#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> g, gr;
vector<bool> used;
vector<int> order, component;
void dfs1(int v) {
  used[v] = true;
  for (size_t i = 0; i < g[v].size(); ++i)
    if (!used[g[v][i]]) dfs1(g[v][i]);
  order.push_back(v);
}
void dfs2(int v) {
  used[v] = true;
  component.push_back(v);
  for (size_t i = 0; i < gr[v].size(); ++i)
    if (!used[gr[v][i]]) dfs2(gr[v][i]);
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  g.resize(n + 5);
  gr.resize(n + 5);
  vector<int> P(n), Q(n);
  for (int i = 0; i < n; i++) cin >> P[i];
  for (int i = 0; i < n; i++) cin >> Q[i];
  for (int i = 0; i + 1 < n; i++) {
    g[P[i]].push_back(P[i + 1]);
    g[Q[i]].push_back(Q[i + 1]);
    gr[P[i + 1]].push_back(P[i]);
    gr[Q[i + 1]].push_back(Q[i]);
  }
  used.resize(n + 1);
  fill(used.begin(), used.end(), false);
  for (int i = 1; i <= n; ++i)
    if (!used[i]) dfs1(i);
  fill(used.begin(), used.end(), false);
  string str(n, '\0');
  char nxt = 'a';
  int use = 0;
  for (int i = 0; i < n; i++) {
    int v = order[n - i - 1];
    if (!used[v]) {
      dfs2(v);
      for (int x : component) str[x - 1] = nxt;
      if (nxt != 'z') nxt++;
      use++;
      component.clear();
    }
  }
  use = min(use, 26);
  if (use < k)
    cout << "NO\n";
  else
    cout << "YES\n" << str << '\n';
  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;
vector<vector<int>> g, gr;
vector<bool> used;
vector<int> order, component;
void dfs1(int v) {
  used[v] = true;
  for (size_t i = 0; i < g[v].size(); ++i)
    if (!used[g[v][i]]) dfs1(g[v][i]);
  order.push_back(v);
}
void dfs2(int v) {
  used[v] = true;
  component.push_back(v);
  for (size_t i = 0; i < gr[v].size(); ++i)
    if (!used[gr[v][i]]) dfs2(gr[v][i]);
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  g.resize(n + 5);
  gr.resize(n + 5);
  vector<int> P(n), Q(n);
  for (int i = 0; i < n; i++) cin >> P[i];
  for (int i = 0; i < n; i++) cin >> Q[i];
  for (int i = 0; i + 1 < n; i++) {
    g[P[i]].push_back(P[i + 1]);
    g[Q[i]].push_back(Q[i + 1]);
    gr[P[i + 1]].push_back(P[i]);
    gr[Q[i + 1]].push_back(Q[i]);
  }
  used.resize(n + 1);
  fill(used.begin(), used.end(), false);
  for (int i = 1; i <= n; ++i)
    if (!used[i]) dfs1(i);
  fill(used.begin(), used.end(), false);
  string str(n, '\0');
  char nxt = 'a';
  int use = 0;
  for (int i = 0; i < n; i++) {
    int v = order[n - i - 1];
    if (!used[v]) {
      dfs2(v);
      for (int x : component) str[x - 1] = nxt;
      if (nxt != 'z') nxt++;
      use++;
      component.clear();
    }
  }
  use = min(use, 26);
  if (use < k)
    cout << "NO\n";
  else
    cout << "YES\n" << str << '\n';
  return 0;
}

```