#include <bits/stdc++.h>
const int MAXN = 2e5 + 5;
int n, k;
struct Edge {
  int nxt, to;
} e[MAXN << 2], e2[MAXN << 2];
int head[MAXN], cnt = 1, head2[MAXN], cnt2 = 1;
inline void add(int u, int v) {
  e[cnt] = {head[u], v};
  head[u] = cnt++;
}
inline void add2(int u, int v) {
  e2[cnt2] = {head2[u], v};
  head2[u] = cnt2++;
}
int num = 0, id[MAXN];
int ru[MAXN];
int ind = 1, dfn[MAXN], low[MAXN];
int stack[MAXN], top = 0;
bool instack[MAXN];
void dfs(int u) {
  dfn[u] = low[u] = ind++;
  stack[top++] = u;
  instack[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (!dfn[v]) {
      dfs(v);
      low[u] = std::min(low[u], low[v]);
    } else if (instack[v])
      low[u] = std::min(low[v], low[u]);
  }
  if (dfn[u] == low[u]) {
    int v;
    num++;
    do {
      v = stack[--top];
      instack[v] = 0;
      id[v] = num;
    } while (u != v);
  }
}
char ans[MAXN];
void bfs() {
  std::queue<int> q;
  for (int u = 1; u <= num; u++)
    if (!ru[u]) q.push(u);
  char x = 'a';
  while (!q.empty()) {
    int u = q.front();
    ans[u] = x;
    if (x != 'z') x++;
    q.pop();
    for (int i = head2[u]; i; i = e2[i].nxt) {
      int v = e2[i].to;
      ru[v]--;
      if (!ru[v]) q.push(v);
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int x = 0; x < 2; x++) {
    int temp;
    scanf("%d", &temp);
    for (int i = 1, j; i < n; i++) {
      scanf("%d", &j);
      add(temp, j);
      temp = j;
    }
    scanf("%d", &temp);
    for (int i = 1, j; i < n; i++) {
      scanf("%d", &j);
      add(temp, j);
      temp = j;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i);
  if (num < k) {
    puts("NO");
    return 0;
  }
  for (int u = 1; u <= n; u++) {
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      if (id[u] != id[v]) add2(id[u], id[v]), ru[id[v]]++;
    }
  }
  puts("YES");
  bfs();
  for (int i = 1; i <= n; i++) {
    putchar(ans[id[i]]);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAXN = 2e5 + 5;
int n, k;
struct Edge {
  int nxt, to;
} e[MAXN << 2], e2[MAXN << 2];
int head[MAXN], cnt = 1, head2[MAXN], cnt2 = 1;
inline void add(int u, int v) {
  e[cnt] = {head[u], v};
  head[u] = cnt++;
}
inline void add2(int u, int v) {
  e2[cnt2] = {head2[u], v};
  head2[u] = cnt2++;
}
int num = 0, id[MAXN];
int ru[MAXN];
int ind = 1, dfn[MAXN], low[MAXN];
int stack[MAXN], top = 0;
bool instack[MAXN];
void dfs(int u) {
  dfn[u] = low[u] = ind++;
  stack[top++] = u;
  instack[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (!dfn[v]) {
      dfs(v);
      low[u] = std::min(low[u], low[v]);
    } else if (instack[v])
      low[u] = std::min(low[v], low[u]);
  }
  if (dfn[u] == low[u]) {
    int v;
    num++;
    do {
      v = stack[--top];
      instack[v] = 0;
      id[v] = num;
    } while (u != v);
  }
}
char ans[MAXN];
void bfs() {
  std::queue<int> q;
  for (int u = 1; u <= num; u++)
    if (!ru[u]) q.push(u);
  char x = 'a';
  while (!q.empty()) {
    int u = q.front();
    ans[u] = x;
    if (x != 'z') x++;
    q.pop();
    for (int i = head2[u]; i; i = e2[i].nxt) {
      int v = e2[i].to;
      ru[v]--;
      if (!ru[v]) q.push(v);
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int x = 0; x < 2; x++) {
    int temp;
    scanf("%d", &temp);
    for (int i = 1, j; i < n; i++) {
      scanf("%d", &j);
      add(temp, j);
      temp = j;
    }
    scanf("%d", &temp);
    for (int i = 1, j; i < n; i++) {
      scanf("%d", &j);
      add(temp, j);
      temp = j;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i);
  if (num < k) {
    puts("NO");
    return 0;
  }
  for (int u = 1; u <= n; u++) {
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      if (id[u] != id[v]) add2(id[u], id[v]), ru[id[v]]++;
    }
  }
  puts("YES");
  bfs();
  for (int i = 1; i <= n; i++) {
    putchar(ans[id[i]]);
  }
  return 0;
}

```