#include <bits/stdc++.h>
using namespace std;
struct edge {
  int to, next;
} e[400005];
bool vis[200005];
int tot = 0, tim = 0, last[200005], a[200005], dfn[200005], low[200005],
    c[200005], sta[200005], s[200005];
inline void adde(int u, int v) {
  e[++tot] = (edge){v, last[u]}, last[u] = tot;
  return;
}
void tarjan(int root) {
  dfn[root] = low[root] = ++tim, sta[++*sta] = root, vis[root] = 1;
  for (register int i = last[root]; i; i = e[i].next) {
    int ot = e[i].to;
    if (!dfn[ot])
      tarjan(ot), low[root] = min(low[root], low[ot]);
    else if (vis[ot])
      low[root] = min(low[root], dfn[ot]);
  }
  if (dfn[root] == low[root]) {
    int _;
    ++*c;
    do {
      _ = sta[*sta], --*sta, vis[_] = 0, c[_] = *c;
    } while (_ ^ root);
  }
  return;
}
int main() {
  int n, k, lst, x;
  scanf("%d%d", &n, &k);
  if (k > 26) return 0 * puts("NO");
  scanf("%d", a + 1);
  for (register int i = 2; i <= n; ++i)
    scanf("%d", a + i), adde(a[i - 1], a[i]);
  scanf("%d", &lst);
  for (register int i = 1; i < n; ++i) scanf("%d", &x), adde(lst, x), lst = x;
  tarjan(a[1]), s[a[1]] = 0;
  for (register int i = 2; i <= n; ++i)
    s[a[i]] = s[a[i - 1]] + (c[a[i - 1]] != c[a[i]]);
  if (*c < k) return 0 * puts("NO");
  puts("YES");
  for (register int i = 1; i <= n; ++i) printf("%c", 'a' + min(s[i], 25));
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
struct edge {
  int to, next;
} e[400005];
bool vis[200005];
int tot = 0, tim = 0, last[200005], a[200005], dfn[200005], low[200005],
    c[200005], sta[200005], s[200005];
inline void adde(int u, int v) {
  e[++tot] = (edge){v, last[u]}, last[u] = tot;
  return;
}
void tarjan(int root) {
  dfn[root] = low[root] = ++tim, sta[++*sta] = root, vis[root] = 1;
  for (register int i = last[root]; i; i = e[i].next) {
    int ot = e[i].to;
    if (!dfn[ot])
      tarjan(ot), low[root] = min(low[root], low[ot]);
    else if (vis[ot])
      low[root] = min(low[root], dfn[ot]);
  }
  if (dfn[root] == low[root]) {
    int _;
    ++*c;
    do {
      _ = sta[*sta], --*sta, vis[_] = 0, c[_] = *c;
    } while (_ ^ root);
  }
  return;
}
int main() {
  int n, k, lst, x;
  scanf("%d%d", &n, &k);
  if (k > 26) return 0 * puts("NO");
  scanf("%d", a + 1);
  for (register int i = 2; i <= n; ++i)
    scanf("%d", a + i), adde(a[i - 1], a[i]);
  scanf("%d", &lst);
  for (register int i = 1; i < n; ++i) scanf("%d", &x), adde(lst, x), lst = x;
  tarjan(a[1]), s[a[1]] = 0;
  for (register int i = 2; i <= n; ++i)
    s[a[i]] = s[a[i - 1]] + (c[a[i - 1]] != c[a[i]]);
  if (*c < k) return 0 * puts("NO");
  puts("YES");
  for (register int i = 1; i <= n; ++i) printf("%c", 'a' + min(s[i], 25));
  return 0;
}

```