#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1);
const long long inf = 1LL << 30;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const int N = 200200;
vector<int> adj[N], SCC_NUM[N];
int ticks = 1, current_scc = 1;
int d[N], low[N];
bool stacked[N];
int a[N], b[N];
stack<int> s;
int p[N];
void tarjan(int u) {
  d[u] = low[u] = ticks++;
  s.push(u);
  stacked[u] = true;
  const vector<int> &out = adj[u];
  for (int k = 0, m = out.size(); k < m; ++k) {
    const int &v = out[k];
    if (d[v] == -1) {
      tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (stacked[v])
      low[u] = min(low[u], low[v]);
  }
  if (d[u] == low[u]) {
    int v;
    do {
      v = s.top();
      s.pop();
      stacked[v] = false;
      SCC_NUM[current_scc].push_back(v);
    } while (u != v);
    current_scc++;
  }
}
int Par(int i) { return p[i] = (i == p[i]) ? i : Par(p[i]); }
bool same(int i, int j) { return Par(i) == Par(j); }
void mrg(int i, int j) {
  if (same(i, j)) return;
  i = Par(i);
  j = Par(j);
  if (rand() & 1) swap(i, j);
  p[i] = p[j];
}
vector<int> adj2[N];
char ans[N];
bool vis[N];
int deg[N];
int mxc = 0;
void DFS(int u, char c) {
  if (vis[u]) return;
  mxc = max(mxc, c - 'a' + 1);
  vis[u] = 1;
  ans[u] = c;
  for (auto v : adj2[u]) DFS(v, min(c + 1, (int)'z'));
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n - 1; i++) {
    adj[a[i]].push_back(a[i + 1]);
    adj[b[i]].push_back(b[i + 1]);
  }
  memset(d, -1, sizeof d);
  memset(low, -1, sizeof low);
  for (int i = 1; i < N; i++) p[i] = i;
  for (int i = 1; i <= n; i++)
    if (d[i] == -1) tarjan(i);
  for (int i = 1; i < current_scc; i++) {
    int lst = -1;
    for (auto it : SCC_NUM[i]) {
      if (lst != -1) mrg(it, lst);
      lst = it;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int u = Par(a[i]), v = Par(a[i + 1]);
    if (u != v) {
      adj2[u].push_back(v);
      deg[v]++;
    }
    u = Par(b[i]), v = Par(b[i + 1]);
    if (u != v) {
      adj2[u].push_back(v);
      deg[v]++;
    }
  }
  vector<int> rt;
  for (int u = 1; u <= n; u++)
    if (!deg[Par(u)]) rt.push_back(Par(u));
  for (auto u : rt) DFS(u, 'a');
  if (mxc < k) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  for (int u = 1; u <= n; u++) printf("%c", ans[Par(u)]);
}

### Prompt

Create a solution in Cpp for 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 long double pi = acos(-1);
const long long inf = 1LL << 30;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const int N = 200200;
vector<int> adj[N], SCC_NUM[N];
int ticks = 1, current_scc = 1;
int d[N], low[N];
bool stacked[N];
int a[N], b[N];
stack<int> s;
int p[N];
void tarjan(int u) {
  d[u] = low[u] = ticks++;
  s.push(u);
  stacked[u] = true;
  const vector<int> &out = adj[u];
  for (int k = 0, m = out.size(); k < m; ++k) {
    const int &v = out[k];
    if (d[v] == -1) {
      tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (stacked[v])
      low[u] = min(low[u], low[v]);
  }
  if (d[u] == low[u]) {
    int v;
    do {
      v = s.top();
      s.pop();
      stacked[v] = false;
      SCC_NUM[current_scc].push_back(v);
    } while (u != v);
    current_scc++;
  }
}
int Par(int i) { return p[i] = (i == p[i]) ? i : Par(p[i]); }
bool same(int i, int j) { return Par(i) == Par(j); }
void mrg(int i, int j) {
  if (same(i, j)) return;
  i = Par(i);
  j = Par(j);
  if (rand() & 1) swap(i, j);
  p[i] = p[j];
}
vector<int> adj2[N];
char ans[N];
bool vis[N];
int deg[N];
int mxc = 0;
void DFS(int u, char c) {
  if (vis[u]) return;
  mxc = max(mxc, c - 'a' + 1);
  vis[u] = 1;
  ans[u] = c;
  for (auto v : adj2[u]) DFS(v, min(c + 1, (int)'z'));
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n - 1; i++) {
    adj[a[i]].push_back(a[i + 1]);
    adj[b[i]].push_back(b[i + 1]);
  }
  memset(d, -1, sizeof d);
  memset(low, -1, sizeof low);
  for (int i = 1; i < N; i++) p[i] = i;
  for (int i = 1; i <= n; i++)
    if (d[i] == -1) tarjan(i);
  for (int i = 1; i < current_scc; i++) {
    int lst = -1;
    for (auto it : SCC_NUM[i]) {
      if (lst != -1) mrg(it, lst);
      lst = it;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int u = Par(a[i]), v = Par(a[i + 1]);
    if (u != v) {
      adj2[u].push_back(v);
      deg[v]++;
    }
    u = Par(b[i]), v = Par(b[i + 1]);
    if (u != v) {
      adj2[u].push_back(v);
      deg[v]++;
    }
  }
  vector<int> rt;
  for (int u = 1; u <= n; u++)
    if (!deg[Par(u)]) rt.push_back(Par(u));
  for (auto u : rt) DFS(u, 'a');
  if (mxc < k) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  for (int u = 1; u <= n; u++) printf("%c", ans[Par(u)]);
}

```