#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
template <typename T>
using Pair = array<T, 2>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  while (cin >> n >> k) {
    vector<int> p(n);
    vector<int> q(n);
    for (int i = 0; i < (int)n; i++) {
      cin >> p[i];
      p[i]--;
    }
    for (int i = 0; i < (int)n; i++) {
      cin >> q[i];
      q[i]--;
    }
    vector<int> nq(n, -1);
    vector<int> np(n, -1);
    for (int i = 0; i < (int)n - 1; i++) {
      np[p[i]] = p[i + 1];
      nq[q[i]] = q[i + 1];
    };
    ;
    vector<pair<int, int>> tout(n);
    int t = 1;
    vector<int> used(n);
    string label(n, 'a');
    function<void(int, char, bool)> dfs = [&](int i, char c, bool calct) {
      if (i == -1) {
        return;
      }
      if (used[i]) {
        return;
      }
      used[i] = true;
      dfs(np[i], c, calct);
      dfs(nq[i], c, calct);
      if (calct) {
        tout[i].first = t++;
        tout[i].second = i;
      }
      label[i] = c;
    };
    for (int i = 0; i < n; i++) {
      if (!used[i]) dfs(i, 'a', true);
      ;
    }
    sort(tout.rbegin(), tout.rend());
    used.assign(n, false);
    np.assign(n, -1);
    nq.assign(n, -1);
    for (int i = 0; i < (int)n - 1; i++) {
      np[p[i + 1]] = p[i];
      nq[q[i + 1]] = q[i];
    }
    int c = 0;
    char sym = 'a';
    ;
    ;
    ;
    for (int i = 0; i < n; i++) {
      if (!used[tout[i].second]) {
        dfs(tout[i].second, sym++, false);
        sym = min(sym, 'z');
        c++;
      };
    }
    auto ans = label;
    sort(ans.begin(), ans.end());
    auto apply = [](string arr, vector<int> perm) {
      auto res = arr;
      for (int i = 0; i < arr.size(); i++) {
        res[perm[i]] = arr[i];
      }
      return res;
    };
    if (c < k) {
      cout << "NO" << endl;
      break;
    }
    cout << "YES" << endl;
    for (int i = 0; i < (int)n; i++) {
      cout << label[i];
    }
    cout << endl;
  }
}

### Prompt

Your challenge is to write 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;
using i64 = long long;
template <typename T>
using Pair = array<T, 2>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  while (cin >> n >> k) {
    vector<int> p(n);
    vector<int> q(n);
    for (int i = 0; i < (int)n; i++) {
      cin >> p[i];
      p[i]--;
    }
    for (int i = 0; i < (int)n; i++) {
      cin >> q[i];
      q[i]--;
    }
    vector<int> nq(n, -1);
    vector<int> np(n, -1);
    for (int i = 0; i < (int)n - 1; i++) {
      np[p[i]] = p[i + 1];
      nq[q[i]] = q[i + 1];
    };
    ;
    vector<pair<int, int>> tout(n);
    int t = 1;
    vector<int> used(n);
    string label(n, 'a');
    function<void(int, char, bool)> dfs = [&](int i, char c, bool calct) {
      if (i == -1) {
        return;
      }
      if (used[i]) {
        return;
      }
      used[i] = true;
      dfs(np[i], c, calct);
      dfs(nq[i], c, calct);
      if (calct) {
        tout[i].first = t++;
        tout[i].second = i;
      }
      label[i] = c;
    };
    for (int i = 0; i < n; i++) {
      if (!used[i]) dfs(i, 'a', true);
      ;
    }
    sort(tout.rbegin(), tout.rend());
    used.assign(n, false);
    np.assign(n, -1);
    nq.assign(n, -1);
    for (int i = 0; i < (int)n - 1; i++) {
      np[p[i + 1]] = p[i];
      nq[q[i + 1]] = q[i];
    }
    int c = 0;
    char sym = 'a';
    ;
    ;
    ;
    for (int i = 0; i < n; i++) {
      if (!used[tout[i].second]) {
        dfs(tout[i].second, sym++, false);
        sym = min(sym, 'z');
        c++;
      };
    }
    auto ans = label;
    sort(ans.begin(), ans.end());
    auto apply = [](string arr, vector<int> perm) {
      auto res = arr;
      for (int i = 0; i < arr.size(); i++) {
        res[perm[i]] = arr[i];
      }
      return res;
    };
    if (c < k) {
      cout << "NO" << endl;
      break;
    }
    cout << "YES" << endl;
    for (int i = 0; i < (int)n; i++) {
      cout << label[i];
    }
    cout << endl;
  }
}

```