#include <bits/stdc++.h>
using namespace std;
map<int, vector<int>> G, GT;
vector<int> order, used;
vector<vector<int>> comp;
void ins(map<int, vector<int>> &g, int x, int y) {
  auto &v = g[x];
  if (find(v.begin(), v.end(), y) == v.end()) {
    v.push_back(y);
  }
}
void dfs1(int x) {
  used[x] = true;
  if (G.count(x))
    for (const int &y : G[x])
      if (!used[y]) dfs1(y);
  order.push_back(x);
}
void dfs2(int x) {
  used[x] = true;
  comp.back().push_back(x);
  if (GT.count(x))
    for (const int &y : GT[x])
      if (!used[y]) dfs2(y);
}
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int _ = 0; _ < 2; ++_)
    for (int x, i = 0, p = 0; i < n; ++i) {
      cin >> x;
      if (p) {
        ins(G, p - 1, x - 1);
        ins(GT, x - 1, p - 1);
      }
      p = x;
    }
  used.assign(n, false);
  for (int i = 0; i < n; ++i)
    if (!used[i]) dfs1(i);
  used.assign(n, false);
  for (int i = 0; i < n; ++i) {
    int v = order[n - 1 - i];
    if (!used[v]) {
      comp.emplace_back();
      dfs2(v);
    }
  }
  if (comp.size() >= k) {
    char c[n + 1];
    c[n] = 0;
    for (int i = 0; i < comp.size(); ++i) {
      for (const int &x : comp[i]) c[x] = 'a' + min('z' - 'a' + 0, i);
    }
    cout << "YES\n" << c << '\n';
  } else
    cout << "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;
map<int, vector<int>> G, GT;
vector<int> order, used;
vector<vector<int>> comp;
void ins(map<int, vector<int>> &g, int x, int y) {
  auto &v = g[x];
  if (find(v.begin(), v.end(), y) == v.end()) {
    v.push_back(y);
  }
}
void dfs1(int x) {
  used[x] = true;
  if (G.count(x))
    for (const int &y : G[x])
      if (!used[y]) dfs1(y);
  order.push_back(x);
}
void dfs2(int x) {
  used[x] = true;
  comp.back().push_back(x);
  if (GT.count(x))
    for (const int &y : GT[x])
      if (!used[y]) dfs2(y);
}
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int _ = 0; _ < 2; ++_)
    for (int x, i = 0, p = 0; i < n; ++i) {
      cin >> x;
      if (p) {
        ins(G, p - 1, x - 1);
        ins(GT, x - 1, p - 1);
      }
      p = x;
    }
  used.assign(n, false);
  for (int i = 0; i < n; ++i)
    if (!used[i]) dfs1(i);
  used.assign(n, false);
  for (int i = 0; i < n; ++i) {
    int v = order[n - 1 - i];
    if (!used[v]) {
      comp.emplace_back();
      dfs2(v);
    }
  }
  if (comp.size() >= k) {
    char c[n + 1];
    c[n] = 0;
    for (int i = 0; i < comp.size(); ++i) {
      for (const int &x : comp[i]) c[x] = 'a' + min('z' - 'a' + 0, i);
    }
    cout << "YES\n" << c << '\n';
  } else
    cout << "NO\n";
  return 0;
}

```