#include <bits/stdc++.h>
const int N = 3e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 1e9;
using namespace std;
int v[N], w[N], mx[N], p[N], sub[N], ans[N];
int root(int u) {
  while (u != p[u]) {
    p[u] = p[p[u]];
    u = p[u];
  }
  return u;
}
void merge(int uu, int vv) {
  int u = root(uu), v = root(vv);
  if (u == v) return;
  if (sub[u] < sub[v]) swap(u, v);
  mx[u] = max(mx[u], mx[v]);
  sub[u] += sub[v];
  p[v] = u;
}
void solve() {
  int n, i, k, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> v[i];
    p[i] = i;
    sub[i] = 1;
    mx[v[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    cin >> w[i];
    mx[w[i]] = max(mx[w[i]], i);
  }
  for (i = 1; i <= n; i++) {
    merge(v[i], w[i]);
  }
  int fill = 0, dist = 0;
  for (i = 1; i <= n; i++) {
    int go = mx[root(v[i])];
    assert(go >= i);
    j = i + 1;
    while (j <= go) {
      merge(v[j], v[i]);
      go = mx[root(v[i])];
      j++;
    }
    for (j = i; j <= go; j++) ans[v[j]] = fill;
    fill = min(fill + 1, k - 1);
    ++dist;
    i = go;
  }
  if (dist < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (i = 1; i <= n; i++) cout << (char)('a' + ans[i]);
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int T = 1;
  for (int t = 1; t <= T; t++) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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>
const int N = 3e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 1e9;
using namespace std;
int v[N], w[N], mx[N], p[N], sub[N], ans[N];
int root(int u) {
  while (u != p[u]) {
    p[u] = p[p[u]];
    u = p[u];
  }
  return u;
}
void merge(int uu, int vv) {
  int u = root(uu), v = root(vv);
  if (u == v) return;
  if (sub[u] < sub[v]) swap(u, v);
  mx[u] = max(mx[u], mx[v]);
  sub[u] += sub[v];
  p[v] = u;
}
void solve() {
  int n, i, k, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> v[i];
    p[i] = i;
    sub[i] = 1;
    mx[v[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    cin >> w[i];
    mx[w[i]] = max(mx[w[i]], i);
  }
  for (i = 1; i <= n; i++) {
    merge(v[i], w[i]);
  }
  int fill = 0, dist = 0;
  for (i = 1; i <= n; i++) {
    int go = mx[root(v[i])];
    assert(go >= i);
    j = i + 1;
    while (j <= go) {
      merge(v[j], v[i]);
      go = mx[root(v[i])];
      j++;
    }
    for (j = i; j <= go; j++) ans[v[j]] = fill;
    fill = min(fill + 1, k - 1);
    ++dist;
    i = go;
  }
  if (dist < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (i = 1; i <= n; i++) cout << (char)('a' + ans[i]);
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int T = 1;
  for (int t = 1; t <= T; t++) {
    solve();
  }
  return 0;
}

```