#include <bits/stdc++.h>
using namespace std;
int n, k, cnt;
int viz1[200005], viz2[200005];
vector<int> G[200005], GR[200005];
stack<int> S;
char l, ans[200005];
void postDFS(int nod) {
  viz1[nod] = 1;
  for (int i = 0; i < G[nod].size(); i++)
    if (!viz1[G[nod][i]]) postDFS(G[nod][i]);
  S.push(nod);
}
void preDFS(int nod) {
  ans[nod] = l;
  viz2[nod] = 1;
  for (int i = 0; i < GR[nod].size(); i++)
    if (!viz2[GR[nod][i]]) preDFS(GR[nod][i]);
}
int main() {
  cin >> n >> k;
  int x, y;
  cin >> x;
  for (int i = 1; i < n; i++) {
    cin >> y;
    G[x].push_back(y);
    GR[y].push_back(x);
    x = y;
  }
  cin >> x;
  for (int i = 1; i < n; i++) {
    cin >> y;
    G[x].push_back(y);
    GR[y].push_back(x);
    x = y;
  }
  for (int i = 1; i <= n; i++)
    if (!viz1[i]) postDFS(i);
  l = 'a';
  while (!S.empty()) {
    int nod = S.top();
    S.pop();
    if (!viz2[nod]) {
      ++cnt;
      preDFS(nod);
      if (cnt < 26) ++l;
    }
  }
  if (cnt < k)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << ans[i];
  }
}

### 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, cnt;
int viz1[200005], viz2[200005];
vector<int> G[200005], GR[200005];
stack<int> S;
char l, ans[200005];
void postDFS(int nod) {
  viz1[nod] = 1;
  for (int i = 0; i < G[nod].size(); i++)
    if (!viz1[G[nod][i]]) postDFS(G[nod][i]);
  S.push(nod);
}
void preDFS(int nod) {
  ans[nod] = l;
  viz2[nod] = 1;
  for (int i = 0; i < GR[nod].size(); i++)
    if (!viz2[GR[nod][i]]) preDFS(GR[nod][i]);
}
int main() {
  cin >> n >> k;
  int x, y;
  cin >> x;
  for (int i = 1; i < n; i++) {
    cin >> y;
    G[x].push_back(y);
    GR[y].push_back(x);
    x = y;
  }
  cin >> x;
  for (int i = 1; i < n; i++) {
    cin >> y;
    G[x].push_back(y);
    GR[y].push_back(x);
    x = y;
  }
  for (int i = 1; i <= n; i++)
    if (!viz1[i]) postDFS(i);
  l = 'a';
  while (!S.empty()) {
    int nod = S.top();
    S.pop();
    if (!viz2[nod]) {
      ++cnt;
      preDFS(nod);
      if (cnt < 26) ++l;
    }
  }
  if (cnt < k)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << ans[i];
  }
}

```