#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, k;
int p[N], q[N];
int adj[N][2], adjout[N][2];
int sz[N], szout[N];
bool visited[N];
stack<int> st;
char ch[N];
int parent[N];
int find(int x) { return parent[x] == -1 ? x : parent[x] = find(parent[x]); }
void dfs1(int i) {
  if (visited[i]) {
    return;
  }
  visited[i] = true;
  for (int j = 0; j < sz[i]; j++) {
    dfs1(adj[i][j]);
  }
  st.push(i);
}
void dfs2(int i, int p) {
  if (visited[i]) {
    return;
  }
  visited[i] = true;
  if (i != p) parent[i] = p;
  for (int j = 0; j < szout[i]; j++) {
    dfs2(adjout[i][j], p);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  fill(parent, parent + n, -1);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    q[i]--;
  }
  for (int i = 1; i < n; i++) {
    adj[p[i - 1]][sz[p[i - 1]]++] = p[i];
    adjout[p[i]][szout[p[i]]++] = p[i - 1];
    adj[q[i - 1]][sz[q[i - 1]]++] = q[i];
    adjout[q[i]][szout[q[i]]++] = q[i - 1];
  }
  for (int i = 0; i < n; i++) {
    dfs1(i);
  }
  fill(visited, visited + n, false);
  int pars = 0;
  while (st.size()) {
    int x = st.top();
    if (!visited[x]) {
      ch[x] = min(pars, k - 1) + 'a';
      pars++;
    }
    dfs2(x, x);
    st.pop();
  }
  if (pars < k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << ch[find(i)];
  }
  cout << endl;
}

### Prompt

Generate 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 = 2e5 + 5;
int n, k;
int p[N], q[N];
int adj[N][2], adjout[N][2];
int sz[N], szout[N];
bool visited[N];
stack<int> st;
char ch[N];
int parent[N];
int find(int x) { return parent[x] == -1 ? x : parent[x] = find(parent[x]); }
void dfs1(int i) {
  if (visited[i]) {
    return;
  }
  visited[i] = true;
  for (int j = 0; j < sz[i]; j++) {
    dfs1(adj[i][j]);
  }
  st.push(i);
}
void dfs2(int i, int p) {
  if (visited[i]) {
    return;
  }
  visited[i] = true;
  if (i != p) parent[i] = p;
  for (int j = 0; j < szout[i]; j++) {
    dfs2(adjout[i][j], p);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  fill(parent, parent + n, -1);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    q[i]--;
  }
  for (int i = 1; i < n; i++) {
    adj[p[i - 1]][sz[p[i - 1]]++] = p[i];
    adjout[p[i]][szout[p[i]]++] = p[i - 1];
    adj[q[i - 1]][sz[q[i - 1]]++] = q[i];
    adjout[q[i]][szout[q[i]]++] = q[i - 1];
  }
  for (int i = 0; i < n; i++) {
    dfs1(i);
  }
  fill(visited, visited + n, false);
  int pars = 0;
  while (st.size()) {
    int x = st.top();
    if (!visited[x]) {
      ch[x] = min(pars, k - 1) + 'a';
      pars++;
    }
    dfs2(x, x);
    st.pop();
  }
  if (pars < k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << ch[find(i)];
  }
  cout << endl;
}

```