#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > ad, scc(200005);
int vis[200005];
int k;
stack<int> st;
vector<pair<int, int> > edge;
void dfs(int s) {
  vis[s] = 1;
  int i, a, x = ad[s].size();
  for (i = 0; i < x; i++) {
    a = ad[s][i];
    if (!vis[a]) dfs(a);
  }
  st.push(s);
  return;
}
void revdfs(int s) {
  vis[s] = 1;
  scc[k].push_back(s);
  int i, a, x = ad[s].size();
  for (i = 0; i < x; i++) {
    a = ad[s][i];
    if (!vis[a]) revdfs(a);
  }
  return;
}
void normal(int n, int m) {
  int i;
  ad.clear();
  ad.resize(n);
  for (i = 0; i < m; i++) ad[edge[i].first].push_back(edge[i].second);
  fill(vis, vis + n, 0);
  for (i = 0; i < n; i++)
    if (!vis[i]) dfs(i);
  return;
}
void ulta(int n, int m) {
  int a, i;
  ad.clear();
  ad.resize(n);
  for (i = 0; i < m; i++) ad[edge[i].second].push_back(edge[i].first);
  fill(vis, vis + n, 0);
  for (i = 0; i < n; i++) {
    a = st.top(), st.pop();
    if (!vis[a]) revdfs(a), k++;
  }
  return;
}
int main() {
  int n, m, a, b, i, j, x, z;
  scanf("%d%d", &n, &z);
  m = 2 * n - 2;
  edge.resize(m);
  vector<int> v(n);
  for (i = 0; i < n; i++) scanf("%d", &v[i]);
  for (i = 0; i < n - 1; i++) {
    a = v[i], b = v[i + 1];
    a--, b--;
    edge[i] = {a, b};
  }
  for (i = 0; i < n; i++) scanf("%d", &v[i]);
  for (i = 0; i < n - 1; i++) {
    a = v[i], b = v[i + 1];
    a--, b--;
    edge[n - 1 + i] = {a, b};
  }
  normal(n, m);
  ulta(n, m);
  string str;
  str.resize(n);
  for (i = 0; i < k; i++) {
    x = scc[i].size();
    b = min(i, z - 1);
    for (j = 0; j < x; j++) a = scc[i][j], str[a] = 'a' + b;
  }
  if (k < z)
    printf("NO\n");
  else {
    printf("YES\n");
    cout << str << endl;
  }
}

### 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>
using namespace std;
vector<vector<int> > ad, scc(200005);
int vis[200005];
int k;
stack<int> st;
vector<pair<int, int> > edge;
void dfs(int s) {
  vis[s] = 1;
  int i, a, x = ad[s].size();
  for (i = 0; i < x; i++) {
    a = ad[s][i];
    if (!vis[a]) dfs(a);
  }
  st.push(s);
  return;
}
void revdfs(int s) {
  vis[s] = 1;
  scc[k].push_back(s);
  int i, a, x = ad[s].size();
  for (i = 0; i < x; i++) {
    a = ad[s][i];
    if (!vis[a]) revdfs(a);
  }
  return;
}
void normal(int n, int m) {
  int i;
  ad.clear();
  ad.resize(n);
  for (i = 0; i < m; i++) ad[edge[i].first].push_back(edge[i].second);
  fill(vis, vis + n, 0);
  for (i = 0; i < n; i++)
    if (!vis[i]) dfs(i);
  return;
}
void ulta(int n, int m) {
  int a, i;
  ad.clear();
  ad.resize(n);
  for (i = 0; i < m; i++) ad[edge[i].second].push_back(edge[i].first);
  fill(vis, vis + n, 0);
  for (i = 0; i < n; i++) {
    a = st.top(), st.pop();
    if (!vis[a]) revdfs(a), k++;
  }
  return;
}
int main() {
  int n, m, a, b, i, j, x, z;
  scanf("%d%d", &n, &z);
  m = 2 * n - 2;
  edge.resize(m);
  vector<int> v(n);
  for (i = 0; i < n; i++) scanf("%d", &v[i]);
  for (i = 0; i < n - 1; i++) {
    a = v[i], b = v[i + 1];
    a--, b--;
    edge[i] = {a, b};
  }
  for (i = 0; i < n; i++) scanf("%d", &v[i]);
  for (i = 0; i < n - 1; i++) {
    a = v[i], b = v[i + 1];
    a--, b--;
    edge[n - 1 + i] = {a, b};
  }
  normal(n, m);
  ulta(n, m);
  string str;
  str.resize(n);
  for (i = 0; i < k; i++) {
    x = scc[i].size();
    b = min(i, z - 1);
    for (j = 0; j < x; j++) a = scc[i][j], str[a] = 'a' + b;
  }
  if (k < z)
    printf("NO\n");
  else {
    printf("YES\n");
    cout << str << endl;
  }
}

```