#include <bits/stdc++.h>
using namespace std;
int p[3][200005], vis[200005], flag[200005], gs, qlt[200005], tim, n, k,
    a[200005], b[200005], t[200005], fa[200005], top, str[200005];
char anss[200005];
void dfs(int x) {
  tim++;
  t[x] = tim;
  fa[x] = t[x];
  vis[x] = 1;
  top++;
  str[top] = x;
  if (p[1][x] + 1 <= n && flag[a[p[1][x] + 1]] == 0) {
    if (t[a[p[1][x] + 1]] > 9999991) {
      dfs(a[p[1][x] + 1]);
      fa[x] = min(fa[x], fa[a[p[1][x] + 1]]);
    } else {
      if (vis[a[p[1][x] + 1]] == 1) fa[x] = min(fa[x], fa[a[p[1][x] + 1]]);
    }
  }
  if (p[2][x] + 1 <= n && flag[b[p[2][x] + 1]] == 0) {
    if (t[b[p[2][x] + 1]] > 9999991) {
      dfs(b[p[2][x] + 1]);
      fa[x] = min(fa[x], fa[b[p[2][x] + 1]]);
    } else {
      if (vis[b[p[2][x] + 1]] == 1) fa[x] = min(fa[x], fa[b[p[2][x] + 1]]);
    }
  }
  if (t[x] == fa[x]) {
    gs++;
    while (t[str[top]] != fa[str[top]]) {
      vis[str[top]] = 0;
      qlt[str[top]] = gs;
      flag[str[top]] = 1;
      top--;
    }
    qlt[str[top]] = gs;
    vis[str[top]] = 0;
    flag[str[top]] = 1;
    top--;
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int tem;
    cin >> tem;
    a[i] = tem;
    p[1][tem] = i;
  }
  for (int i = 1; i <= n; i++) {
    int tem;
    cin >> tem;
    b[i] = tem;
    p[2][tem] = i;
  }
  for (int i = 1; i <= n; i++) {
    t[i] = 9999999;
  }
  dfs(a[1]);
  for (int i = 1; i <= n; i++) {
    if (gs - qlt[i] > 25)
      anss[i] = 'z';
    else
      anss[i] = gs - qlt[i] + 'a';
  }
  if (gs >= k) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << anss[i];
  } else
    cout << "NO";
}

### 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;
int p[3][200005], vis[200005], flag[200005], gs, qlt[200005], tim, n, k,
    a[200005], b[200005], t[200005], fa[200005], top, str[200005];
char anss[200005];
void dfs(int x) {
  tim++;
  t[x] = tim;
  fa[x] = t[x];
  vis[x] = 1;
  top++;
  str[top] = x;
  if (p[1][x] + 1 <= n && flag[a[p[1][x] + 1]] == 0) {
    if (t[a[p[1][x] + 1]] > 9999991) {
      dfs(a[p[1][x] + 1]);
      fa[x] = min(fa[x], fa[a[p[1][x] + 1]]);
    } else {
      if (vis[a[p[1][x] + 1]] == 1) fa[x] = min(fa[x], fa[a[p[1][x] + 1]]);
    }
  }
  if (p[2][x] + 1 <= n && flag[b[p[2][x] + 1]] == 0) {
    if (t[b[p[2][x] + 1]] > 9999991) {
      dfs(b[p[2][x] + 1]);
      fa[x] = min(fa[x], fa[b[p[2][x] + 1]]);
    } else {
      if (vis[b[p[2][x] + 1]] == 1) fa[x] = min(fa[x], fa[b[p[2][x] + 1]]);
    }
  }
  if (t[x] == fa[x]) {
    gs++;
    while (t[str[top]] != fa[str[top]]) {
      vis[str[top]] = 0;
      qlt[str[top]] = gs;
      flag[str[top]] = 1;
      top--;
    }
    qlt[str[top]] = gs;
    vis[str[top]] = 0;
    flag[str[top]] = 1;
    top--;
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int tem;
    cin >> tem;
    a[i] = tem;
    p[1][tem] = i;
  }
  for (int i = 1; i <= n; i++) {
    int tem;
    cin >> tem;
    b[i] = tem;
    p[2][tem] = i;
  }
  for (int i = 1; i <= n; i++) {
    t[i] = 9999999;
  }
  dfs(a[1]);
  for (int i = 1; i <= n; i++) {
    if (gs - qlt[i] > 25)
      anss[i] = 'z';
    else
      anss[i] = gs - qlt[i] + 'a';
  }
  if (gs >= k) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << anss[i];
  } else
    cout << "NO";
}

```