#include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
const int N = 2e5 + 5;
int n, k, a[N];
int tplt[N], dem, deg[N], color[N];
int low[N], num[N], dd[N], cnt;
vector<int> adj[N], dsk[N];
stack<int> stk;
void dfs(int u) {
  stk.push(u);
  low[u] = num[u] = ++cnt;
  for (int v : adj[u])
    if (!dd[v]) {
      if (num[v])
        low[u] = min(low[u], low[v]);
      else {
        dfs(v);
        low[u] = min(low[u], low[v]);
      }
    }
  if (low[u] == num[u]) {
    int v;
    ++dem;
    do {
      v = stk.top();
      stk.pop();
      dd[v] = 1;
      tplt[v] = dem;
    } while (v != u);
  }
}
void gogo() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i >= 2) {
      adj[a[i - 1]].push_back(a[i]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i >= 2) {
      adj[a[i - 1]].push_back(a[i]);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!dd[i]) {
      dfs(i);
    }
  for (int u = 1; u <= n; ++u) {
    for (int v : adj[u]) {
      if (tplt[u] != tplt[v]) {
        dsk[tplt[u]].push_back(tplt[v]);
      }
    }
  }
  for (int i = 1; i <= dem; ++i) {
    for (int j : dsk[i]) {
      ++deg[j];
    }
  }
  deque<int> q;
  for (int i = 1; i <= dem; ++i) {
    if (!deg[i]) {
      q.push_back(i);
    }
  }
  int c = -1;
  while (((int)(q).size())) {
    int u = q.front();
    q.pop_front();
    ++c;
    c = min(c, 25);
    color[u] = c;
    for (int v : dsk[u]) {
      if (--deg[v] == 0) {
        q.push_back(v);
      }
    }
  }
  if (c + 1 < k) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << char(color[tplt[i]] + 'a');
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  gogo();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
const int N = 2e5 + 5;
int n, k, a[N];
int tplt[N], dem, deg[N], color[N];
int low[N], num[N], dd[N], cnt;
vector<int> adj[N], dsk[N];
stack<int> stk;
void dfs(int u) {
  stk.push(u);
  low[u] = num[u] = ++cnt;
  for (int v : adj[u])
    if (!dd[v]) {
      if (num[v])
        low[u] = min(low[u], low[v]);
      else {
        dfs(v);
        low[u] = min(low[u], low[v]);
      }
    }
  if (low[u] == num[u]) {
    int v;
    ++dem;
    do {
      v = stk.top();
      stk.pop();
      dd[v] = 1;
      tplt[v] = dem;
    } while (v != u);
  }
}
void gogo() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i >= 2) {
      adj[a[i - 1]].push_back(a[i]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i >= 2) {
      adj[a[i - 1]].push_back(a[i]);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!dd[i]) {
      dfs(i);
    }
  for (int u = 1; u <= n; ++u) {
    for (int v : adj[u]) {
      if (tplt[u] != tplt[v]) {
        dsk[tplt[u]].push_back(tplt[v]);
      }
    }
  }
  for (int i = 1; i <= dem; ++i) {
    for (int j : dsk[i]) {
      ++deg[j];
    }
  }
  deque<int> q;
  for (int i = 1; i <= dem; ++i) {
    if (!deg[i]) {
      q.push_back(i);
    }
  }
  int c = -1;
  while (((int)(q).size())) {
    int u = q.front();
    q.pop_front();
    ++c;
    c = min(c, 25);
    color[u] = c;
    for (int v : dsk[u]) {
      if (--deg[v] == 0) {
        q.push_back(v);
      }
    }
  }
  if (c + 1 < k) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << char(color[tplt[i]] + 'a');
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  gogo();
  return 0;
}

```