#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
stack<int> st;
vector<int> adj[N], B[N], A[N], rev[N];
queue<int> q;
int vis[N], visit[N], a[N], b[N], in[N];
string s[N];
map<int, int> mp[N];
void dfs(int r) {
  vis[r] = 1;
  for (auto it : adj[r])
    if (!vis[it]) dfs(it);
  st.push(r);
}
void dfs(int r, int p) {
  vis[r] = p;
  if (r != p) B[p].push_back(r);
  for (auto it : rev[r])
    if (!vis[it]) dfs(it, p);
}
void run() {
  int n, k, x, i;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) cin >> b[i];
  for (i = 1; i < n; i++)
    adj[a[i]].push_back(a[i + 1]), adj[b[i]].push_back(b[i + 1]),
        rev[a[i + 1]].push_back(a[i]), rev[b[i + 1]].push_back(b[i]);
  for (i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
  }
  memset(vis, 0, sizeof vis);
  while (!st.empty()) {
    x = st.top();
    st.pop();
    if (vis[x]) continue;
    dfs(x, x);
  }
  for (i = 1; i <= n; i++) {
    for (auto it : adj[i]) {
      if (!mp[vis[i]][vis[it]] && vis[i] != vis[it]) {
        A[vis[i]].push_back(vis[it]);
        in[vis[it]]++;
        mp[vis[i]][vis[it]] = 1;
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (!in[i] && vis[i] == i) {
      visit[i] = 1;
      q.push(i);
    }
  }
  vector<int> ans;
  while (!q.empty()) {
    x = q.front();
    ans.push_back(x);
    q.pop();
    for (auto it : A[x]) {
      if (!visit[it]) {
        in[it]--;
        if (!in[it]) {
          visit[it] = 1;
          q.push(it);
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (vis[i] == i && visit[i] == 0) {
      cout << "NO";
      return;
    }
  }
  x = (int)ans.size();
  if (x < k) {
    cout << "NO";
    return;
  }
  for (i = 0; i < k; i++) {
    s[ans[i]] = (char)(i + 'a');
    for (auto it : B[ans[i]]) s[it] = s[ans[i]];
  }
  for (i = k; i < x; i++) {
    s[ans[i]] = 'z';
    for (auto it : B[ans[i]]) s[it] = 'z';
  }
  cout << "YES\n";
  for (i = 1; i <= n; i++) cout << s[i];
}
void setup() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cout << setprecision(20);
}
signed main() {
  setup();
  int tests = 1;
  for (int i = 1; i <= tests; i++) run();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 = 200005;
stack<int> st;
vector<int> adj[N], B[N], A[N], rev[N];
queue<int> q;
int vis[N], visit[N], a[N], b[N], in[N];
string s[N];
map<int, int> mp[N];
void dfs(int r) {
  vis[r] = 1;
  for (auto it : adj[r])
    if (!vis[it]) dfs(it);
  st.push(r);
}
void dfs(int r, int p) {
  vis[r] = p;
  if (r != p) B[p].push_back(r);
  for (auto it : rev[r])
    if (!vis[it]) dfs(it, p);
}
void run() {
  int n, k, x, i;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) cin >> b[i];
  for (i = 1; i < n; i++)
    adj[a[i]].push_back(a[i + 1]), adj[b[i]].push_back(b[i + 1]),
        rev[a[i + 1]].push_back(a[i]), rev[b[i + 1]].push_back(b[i]);
  for (i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
  }
  memset(vis, 0, sizeof vis);
  while (!st.empty()) {
    x = st.top();
    st.pop();
    if (vis[x]) continue;
    dfs(x, x);
  }
  for (i = 1; i <= n; i++) {
    for (auto it : adj[i]) {
      if (!mp[vis[i]][vis[it]] && vis[i] != vis[it]) {
        A[vis[i]].push_back(vis[it]);
        in[vis[it]]++;
        mp[vis[i]][vis[it]] = 1;
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (!in[i] && vis[i] == i) {
      visit[i] = 1;
      q.push(i);
    }
  }
  vector<int> ans;
  while (!q.empty()) {
    x = q.front();
    ans.push_back(x);
    q.pop();
    for (auto it : A[x]) {
      if (!visit[it]) {
        in[it]--;
        if (!in[it]) {
          visit[it] = 1;
          q.push(it);
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (vis[i] == i && visit[i] == 0) {
      cout << "NO";
      return;
    }
  }
  x = (int)ans.size();
  if (x < k) {
    cout << "NO";
    return;
  }
  for (i = 0; i < k; i++) {
    s[ans[i]] = (char)(i + 'a');
    for (auto it : B[ans[i]]) s[it] = s[ans[i]];
  }
  for (i = k; i < x; i++) {
    s[ans[i]] = 'z';
    for (auto it : B[ans[i]]) s[it] = 'z';
  }
  cout << "YES\n";
  for (i = 1; i <= n; i++) cout << s[i];
}
void setup() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cout << setprecision(20);
}
signed main() {
  setup();
  int tests = 1;
  for (int i = 1; i <= tests; i++) run();
  return 0;
}

```