#include <bits/stdc++.h>
using namespace std;
struct edge {
  int des, next;
} e[200005 << 3];
int n;
int m;
int a[200005];
queue<int> q;
int cnt;
int head[200005];
int newhead[200005];
void ins(int u, int v, int head[]) {
  ++cnt;
  e[cnt].des = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
int dfn[200005];
int low[200005];
int tim;
int in[200005];
stack<int> stk;
int sccid[200005];
int sccnum;
void tarjan(int u) {
  dfn[u] = low[u] = ++tim;
  in[u] = true;
  stk.push(u);
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].des;
    if (!dfn[v]) {
      tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (in[v])
      low[u] = min(low[u], low[v]);
  }
  if (dfn[u] == low[u]) {
    ++sccnum;
    int tmp;
    do {
      tmp = stk.top();
      stk.pop();
      in[tmp] = false;
      sccid[tmp] = sccnum;
    } while (u != tmp);
  }
}
int rd[200005];
char ans[200005];
int vis[30];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i < n; ++i) ins(a[i], a[i + 1], head);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i < n; ++i) ins(a[i], a[i + 1], head);
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) tarjan(i);
  for (int i = 1; i <= n; ++i) {
    int u = sccid[i];
    for (int j = head[i]; j; j = e[j].next) {
      int v = sccid[e[j].des];
      if (u == v) continue;
      ins(u, v, newhead);
      ++rd[v];
    }
  }
  char c = 'a';
  for (int i = 1; i <= sccnum; ++i)
    if (!rd[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    ans[u] = c;
    if (c != 'z') ++c;
    for (int i = newhead[u]; i; i = e[i].next) {
      int v = e[i].des;
      if (--rd[v]) q.push(v);
    }
  }
  if (sccnum < m)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= n; ++i) printf("%c", ans[sccid[i]]);
    printf("\n");
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
struct edge {
  int des, next;
} e[200005 << 3];
int n;
int m;
int a[200005];
queue<int> q;
int cnt;
int head[200005];
int newhead[200005];
void ins(int u, int v, int head[]) {
  ++cnt;
  e[cnt].des = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
int dfn[200005];
int low[200005];
int tim;
int in[200005];
stack<int> stk;
int sccid[200005];
int sccnum;
void tarjan(int u) {
  dfn[u] = low[u] = ++tim;
  in[u] = true;
  stk.push(u);
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].des;
    if (!dfn[v]) {
      tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (in[v])
      low[u] = min(low[u], low[v]);
  }
  if (dfn[u] == low[u]) {
    ++sccnum;
    int tmp;
    do {
      tmp = stk.top();
      stk.pop();
      in[tmp] = false;
      sccid[tmp] = sccnum;
    } while (u != tmp);
  }
}
int rd[200005];
char ans[200005];
int vis[30];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i < n; ++i) ins(a[i], a[i + 1], head);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i < n; ++i) ins(a[i], a[i + 1], head);
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) tarjan(i);
  for (int i = 1; i <= n; ++i) {
    int u = sccid[i];
    for (int j = head[i]; j; j = e[j].next) {
      int v = sccid[e[j].des];
      if (u == v) continue;
      ins(u, v, newhead);
      ++rd[v];
    }
  }
  char c = 'a';
  for (int i = 1; i <= sccnum; ++i)
    if (!rd[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    ans[u] = c;
    if (c != 'z') ++c;
    for (int i = newhead[u]; i; i = e[i].next) {
      int v = e[i].des;
      if (--rd[v]) q.push(v);
    }
  }
  if (sccnum < m)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= n; ++i) printf("%c", ans[sccid[i]]);
    printf("\n");
  }
  return 0;
}

```