#include <bits/stdc++.h>
using namespace std;
int n, ind[200007], low[200007], onstack[200007], last, n2, t[200007];
vector<long long> adj[200007], st;
void go(int v) {
  ind[v] = low[v] = ++last;
  st.push_back(v);
  onstack[v] = 1;
  for (auto w : adj[v]) {
    if (!ind[w]) {
      go(w);
      low[v] = min(low[v], low[w]);
    } else if (onstack[w]) {
      low[v] = min(low[v], low[w]);
    }
  }
  if (low[v] == ind[v]) {
    ++n2;
    for (int w = -1; w != v;) {
      w = st.back();
      st.pop_back();
      onstack[w] = 0;
      t[w] = n2;
    }
  }
}
void scc() {
  last = n2 = 0;
  for (int i = 1; i < n + 1; i++) ind[i] = onstack[i] = 0;
  for (int i = 1; i < n + 1; i++)
    if (!ind[i]) go(i);
  for (int i = 1; i < n + 1; i++) t[i] = n2 + 1 - t[i];
}
vector<long long> cc[200007];
vector<long long> gg[200007];
stack<int> order;
int vis[200005];
bool dfs(int node) {
  vis[node] = 1;
  bool b = 0;
  for (int i = 0; i < (int)gg[node].size(); i++) {
    if (vis[gg[node][i]] == 0) {
      b = b || dfs(gg[node][i]);
    } else if (vis[node] == 1) {
      b = 1;
      break;
    }
  }
  vis[node] = 2;
  order.push(node);
  return b;
}
char ma[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) adj[a[i]].push_back(a[i + 1]);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n + 1; i++) {
    sort(adj[i].begin(), adj[i].end());
  }
  for (int i = 0; i < n - 1; i++) {
    if (!binary_search(adj[a[i]].begin(), adj[a[i]].end(), a[i + 1]))
      adj[a[i]].push_back(a[i + 1]);
  }
  scc();
  for (int i = 1; i < n + 1; i++) {
    cc[t[i]].push_back(i);
  }
  long long x = 0;
  for (int i = 1; i < n + 1; i++) x += (int((cc[i]).size()) > 0);
  if (x < k)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    int off = 0;
    for (int i = 1; i < n + 1; i++) {
      for (int j = 0; j < int((cc[i]).size()); j++) {
        ma[cc[i][j]] = 'a' + off;
      }
      off = min(off + 1, (int)k - 1);
    }
    for (int i = 1; i < n + 1; i++) {
      assert(ma[i] >= 'a');
      cout << ma[i];
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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, ind[200007], low[200007], onstack[200007], last, n2, t[200007];
vector<long long> adj[200007], st;
void go(int v) {
  ind[v] = low[v] = ++last;
  st.push_back(v);
  onstack[v] = 1;
  for (auto w : adj[v]) {
    if (!ind[w]) {
      go(w);
      low[v] = min(low[v], low[w]);
    } else if (onstack[w]) {
      low[v] = min(low[v], low[w]);
    }
  }
  if (low[v] == ind[v]) {
    ++n2;
    for (int w = -1; w != v;) {
      w = st.back();
      st.pop_back();
      onstack[w] = 0;
      t[w] = n2;
    }
  }
}
void scc() {
  last = n2 = 0;
  for (int i = 1; i < n + 1; i++) ind[i] = onstack[i] = 0;
  for (int i = 1; i < n + 1; i++)
    if (!ind[i]) go(i);
  for (int i = 1; i < n + 1; i++) t[i] = n2 + 1 - t[i];
}
vector<long long> cc[200007];
vector<long long> gg[200007];
stack<int> order;
int vis[200005];
bool dfs(int node) {
  vis[node] = 1;
  bool b = 0;
  for (int i = 0; i < (int)gg[node].size(); i++) {
    if (vis[gg[node][i]] == 0) {
      b = b || dfs(gg[node][i]);
    } else if (vis[node] == 1) {
      b = 1;
      break;
    }
  }
  vis[node] = 2;
  order.push(node);
  return b;
}
char ma[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) adj[a[i]].push_back(a[i + 1]);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n + 1; i++) {
    sort(adj[i].begin(), adj[i].end());
  }
  for (int i = 0; i < n - 1; i++) {
    if (!binary_search(adj[a[i]].begin(), adj[a[i]].end(), a[i + 1]))
      adj[a[i]].push_back(a[i + 1]);
  }
  scc();
  for (int i = 1; i < n + 1; i++) {
    cc[t[i]].push_back(i);
  }
  long long x = 0;
  for (int i = 1; i < n + 1; i++) x += (int((cc[i]).size()) > 0);
  if (x < k)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    int off = 0;
    for (int i = 1; i < n + 1; i++) {
      for (int j = 0; j < int((cc[i]).size()); j++) {
        ma[cc[i][j]] = 'a' + off;
      }
      off = min(off + 1, (int)k - 1);
    }
    for (int i = 1; i < n + 1; i++) {
      assert(ma[i] >= 'a');
      cout << ma[i];
    }
    cout << endl;
  }
  return 0;
}

```