#include <bits/stdc++.h>
const int N = 2e5 + 52;
struct edge {
  int to, next;
} e[N << 1];
int head[N], cnt, deg[N];
void addedge(int x, int y, int f = 0) {
  e[++cnt] = (edge){y, x[head]}, x[head] = cnt, y[deg] += f;
}
int dfn[N], low[N], ins[N], col[N], scnt, idx;
std::vector<int> scc[N];
std::stack<int> st;
void dfs(int x) {
  x[dfn] = x[low] = ++idx, x[ins] = 1, st.push(x);
  for (int i = x[head], nx; i; i = e[i].next)
    if (!(nx = e[i].to)[dfn])
      dfs(nx), x[low] = std::min(x[low], nx[low]);
    else if (nx[ins])
      x[low] = std::min(x[low], nx[dfn]);
  if (x[dfn] == x[low])
    for (int y = scnt++; !st.empty();) {
      y = st.top(), st.pop(), scc[scnt].push_back(y), y[col] = scnt, y[ins] = 0;
      if (y == x) break;
    }
}
int n, k, a[N], b[N];
std::vector<int> ord;
std::queue<int> q;
char ans[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  for (int i = 1; i < n; i++) addedge(a[i], a[i + 1]), addedge(b[i], b[i + 1]);
  for (int i = 1; i <= n; i++)
    if (!i[dfn]) dfs(i);
  memset(head, 0, sizeof head), cnt = 0;
  for (int i = 1; i < n; i++) {
    if (a[i][col] != a[i + 1][col]) addedge(a[i][col], a[i + 1][col], 1);
    if (b[i][col] != b[i + 1][col]) addedge(b[i][col], b[i + 1][col], 1);
  }
  for (int i = 1; i <= scnt; i++)
    if (!i[deg]) q.push(i);
  for (; !q.empty(); q.pop())
    for (int x = q.front(), i = x[head], nx = (ord.push_back(x), 1); i;
         i = e[i].next)
      if (!--(nx = e[i].to)[deg]) q.push(nx);
  if (ord.size() < k) return puts("NO"), 0;
  for (int i = 0; i < ord.size(); i++)
    for (int j : ord[i][scc]) ans[j] = std::min(122, 97 + i);
  puts("YES");
  puts(ans + 1);
}

### Prompt

Your task is to create 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>
const int N = 2e5 + 52;
struct edge {
  int to, next;
} e[N << 1];
int head[N], cnt, deg[N];
void addedge(int x, int y, int f = 0) {
  e[++cnt] = (edge){y, x[head]}, x[head] = cnt, y[deg] += f;
}
int dfn[N], low[N], ins[N], col[N], scnt, idx;
std::vector<int> scc[N];
std::stack<int> st;
void dfs(int x) {
  x[dfn] = x[low] = ++idx, x[ins] = 1, st.push(x);
  for (int i = x[head], nx; i; i = e[i].next)
    if (!(nx = e[i].to)[dfn])
      dfs(nx), x[low] = std::min(x[low], nx[low]);
    else if (nx[ins])
      x[low] = std::min(x[low], nx[dfn]);
  if (x[dfn] == x[low])
    for (int y = scnt++; !st.empty();) {
      y = st.top(), st.pop(), scc[scnt].push_back(y), y[col] = scnt, y[ins] = 0;
      if (y == x) break;
    }
}
int n, k, a[N], b[N];
std::vector<int> ord;
std::queue<int> q;
char ans[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  for (int i = 1; i < n; i++) addedge(a[i], a[i + 1]), addedge(b[i], b[i + 1]);
  for (int i = 1; i <= n; i++)
    if (!i[dfn]) dfs(i);
  memset(head, 0, sizeof head), cnt = 0;
  for (int i = 1; i < n; i++) {
    if (a[i][col] != a[i + 1][col]) addedge(a[i][col], a[i + 1][col], 1);
    if (b[i][col] != b[i + 1][col]) addedge(b[i][col], b[i + 1][col], 1);
  }
  for (int i = 1; i <= scnt; i++)
    if (!i[deg]) q.push(i);
  for (; !q.empty(); q.pop())
    for (int x = q.front(), i = x[head], nx = (ord.push_back(x), 1); i;
         i = e[i].next)
      if (!--(nx = e[i].to)[deg]) q.push(nx);
  if (ord.size() < k) return puts("NO"), 0;
  for (int i = 0; i < ord.size(); i++)
    for (int j : ord[i][scc]) ans[j] = std::min(122, 97 + i);
  puts("YES");
  puts(ans + 1);
}

```