#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
void write(int x) {
  printf("%d\n", x);
  return;
}
int n, k, p[N], q[N], deg[N];
vector<int> v[N], tr[N];
int dfn[N], low[N], tim;
bool inq[N];
int col[N], id[N], tot;
int stk[N], t;
char ans[N];
void tarjan(int s) {
  dfn[s] = low[s] = ++tim;
  stk[++t] = s;
  inq[s] = true;
  for (auto t : v[s]) {
    if (!dfn[t]) {
      tarjan(t);
      low[s] = min(low[s], low[t]);
    } else if (inq[t])
      low[s] = min(low[s], dfn[t]);
  }
  if (dfn[s] == low[s]) {
    int cur;
    tot++;
    do {
      cur = stk[t--];
      inq[cur] = false;
      col[cur] = tot;
    } while (cur != s);
  }
}
void tpsort() {
  queue<int> q;
  for (int i = 1; i <= tot; ++i)
    if (deg[i] == 0) q.push(i);
  char base = 'a';
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    ans[s] = base;
    if (base != 'z') ++base;
    for (auto t : tr[s]) {
      deg[t]--;
      if (deg[t] == 0) {
        q.push(t);
      }
    }
  }
}
void main_solve() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) p[i] = read();
  for (int i = 1; i <= n; ++i) q[i] = read();
  for (int i = 1; i < n; ++i) {
    v[p[i]].push_back(p[i + 1]);
    v[q[i]].push_back(q[i + 1]);
  }
  for (int i = 1; i <= n; ++i) {
    if (!dfn[i]) tarjan(i);
  }
  if (tot < k) {
    puts("NO");
    return;
  }
  for (int i = 1; i <= n; ++i) {
    for (auto t : v[i]) {
      if (col[i] != col[t]) {
        tr[col[i]].push_back(col[t]);
        deg[col[t]]++;
      }
    }
  }
  puts("YES");
  tpsort();
  for (int i = 1; i <= n; ++i) {
    printf("%c", ans[col[i]]);
  }
  printf("\n");
  return;
}
signed main() {
  int T = 1;
  while (T--) main_solve();
  return 0;
}

### Prompt

Please formulate 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;
const int N = 2e5 + 5;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
void write(int x) {
  printf("%d\n", x);
  return;
}
int n, k, p[N], q[N], deg[N];
vector<int> v[N], tr[N];
int dfn[N], low[N], tim;
bool inq[N];
int col[N], id[N], tot;
int stk[N], t;
char ans[N];
void tarjan(int s) {
  dfn[s] = low[s] = ++tim;
  stk[++t] = s;
  inq[s] = true;
  for (auto t : v[s]) {
    if (!dfn[t]) {
      tarjan(t);
      low[s] = min(low[s], low[t]);
    } else if (inq[t])
      low[s] = min(low[s], dfn[t]);
  }
  if (dfn[s] == low[s]) {
    int cur;
    tot++;
    do {
      cur = stk[t--];
      inq[cur] = false;
      col[cur] = tot;
    } while (cur != s);
  }
}
void tpsort() {
  queue<int> q;
  for (int i = 1; i <= tot; ++i)
    if (deg[i] == 0) q.push(i);
  char base = 'a';
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    ans[s] = base;
    if (base != 'z') ++base;
    for (auto t : tr[s]) {
      deg[t]--;
      if (deg[t] == 0) {
        q.push(t);
      }
    }
  }
}
void main_solve() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) p[i] = read();
  for (int i = 1; i <= n; ++i) q[i] = read();
  for (int i = 1; i < n; ++i) {
    v[p[i]].push_back(p[i + 1]);
    v[q[i]].push_back(q[i + 1]);
  }
  for (int i = 1; i <= n; ++i) {
    if (!dfn[i]) tarjan(i);
  }
  if (tot < k) {
    puts("NO");
    return;
  }
  for (int i = 1; i <= n; ++i) {
    for (auto t : v[i]) {
      if (col[i] != col[t]) {
        tr[col[i]].push_back(col[t]);
        deg[col[t]]++;
      }
    }
  }
  puts("YES");
  tpsort();
  for (int i = 1; i <= n; ++i) {
    printf("%c", ans[col[i]]);
  }
  printf("\n");
  return;
}
signed main() {
  int T = 1;
  while (T--) main_solve();
  return 0;
}

```