#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-6;
const int INF = 0x3fffffff;
const long long LINF = INF * 1ll * INF;
const double PI = acos(-1.0);
class SCC {
 public:
  int head[200005];
  struct Edge {
    int to, next;
    Edge() {}
    Edge(int to, int next) : to(to), next(next) {}
  } e[200005 << 1];
  void addEdge(int u, int v) {
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt++;
  }
  int belong[200005];
  int dfn[200005];
  int low[200005];
  int ind[200005];
  int outd[200005];
  stack<int> sta;
  vector<int> nG[200005];
  int n, k;
  int cnt;
  int dfs_clock, scc_cnt;
  char ans[200005];
  void init(int n, int k) {
    this->n = n;
    this->k = k;
    dfs_clock = scc_cnt = 0;
    memset(dfn, 0, sizeof(dfn));
    memset(belong, 0, sizeof(belong));
  }
  void tarjan(int u) {
    dfn[u] = low[u] = ++dfs_clock;
    sta.push(u);
    for (int i = head[u]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (!dfn[v]) {
        tarjan(v);
        low[u] = min(low[u], low[v]);
      } else if (!belong[v])
        low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
      scc_cnt++;
      while (!sta.empty()) {
        int temp = sta.top();
        belong[temp] = scc_cnt;
        sta.pop();
        if (u == temp) break;
      }
    }
  }
  void get_scc() {
    memset(ind, 0, sizeof(ind));
    memset(outd, 0, sizeof(outd));
    for (int i = 1; i <= n; i++) {
      for (int j = head[i]; ~j; j = e[j].next) {
        int v = e[j].to;
        if (belong[i] != belong[v]) {
          ind[belong[v]]++;
          outd[belong[i]]++;
          nG[belong[i]].push_back(belong[v]);
        }
      }
    }
  }
  void topoSort() {
    queue<int> q;
    for (int i = 1; i <= scc_cnt; i++) {
      if (!ind[i]) q.push(i);
    }
    int id = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      ans[u] = id + 'a';
      if (id < 25) id++;
      for (int i = 0; i < nG[u].size(); i++) {
        int v = nG[u][i];
        ind[v]--;
        if (!ind[v]) q.push(v);
      }
    }
  }
  void buildGraph() {
    cnt = 0;
    memset(head, -1, sizeof(head));
    int u, v;
    scanf("%d", &u);
    for (int i = 1; i < n; i++) {
      scanf("%d", &v);
      addEdge(u, v);
      u = v;
    }
    scanf("%d", &u);
    for (int i = 1; i < n; i++) {
      scanf("%d", &v);
      addEdge(u, v);
      u = v;
    }
  }
  void solve() {
    buildGraph();
    for (int i = 1; i <= n; i++) {
      if (!dfn[i]) tarjan(i);
    }
    if (scc_cnt < k) {
      puts("NO");
      return;
    }
    get_scc();
    topoSort();
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%c", ans[belong[i]]);
    puts("");
  }
} scc;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  scc.init(n, k);
  scc.solve();
  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;
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-6;
const int INF = 0x3fffffff;
const long long LINF = INF * 1ll * INF;
const double PI = acos(-1.0);
class SCC {
 public:
  int head[200005];
  struct Edge {
    int to, next;
    Edge() {}
    Edge(int to, int next) : to(to), next(next) {}
  } e[200005 << 1];
  void addEdge(int u, int v) {
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt++;
  }
  int belong[200005];
  int dfn[200005];
  int low[200005];
  int ind[200005];
  int outd[200005];
  stack<int> sta;
  vector<int> nG[200005];
  int n, k;
  int cnt;
  int dfs_clock, scc_cnt;
  char ans[200005];
  void init(int n, int k) {
    this->n = n;
    this->k = k;
    dfs_clock = scc_cnt = 0;
    memset(dfn, 0, sizeof(dfn));
    memset(belong, 0, sizeof(belong));
  }
  void tarjan(int u) {
    dfn[u] = low[u] = ++dfs_clock;
    sta.push(u);
    for (int i = head[u]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (!dfn[v]) {
        tarjan(v);
        low[u] = min(low[u], low[v]);
      } else if (!belong[v])
        low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
      scc_cnt++;
      while (!sta.empty()) {
        int temp = sta.top();
        belong[temp] = scc_cnt;
        sta.pop();
        if (u == temp) break;
      }
    }
  }
  void get_scc() {
    memset(ind, 0, sizeof(ind));
    memset(outd, 0, sizeof(outd));
    for (int i = 1; i <= n; i++) {
      for (int j = head[i]; ~j; j = e[j].next) {
        int v = e[j].to;
        if (belong[i] != belong[v]) {
          ind[belong[v]]++;
          outd[belong[i]]++;
          nG[belong[i]].push_back(belong[v]);
        }
      }
    }
  }
  void topoSort() {
    queue<int> q;
    for (int i = 1; i <= scc_cnt; i++) {
      if (!ind[i]) q.push(i);
    }
    int id = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      ans[u] = id + 'a';
      if (id < 25) id++;
      for (int i = 0; i < nG[u].size(); i++) {
        int v = nG[u][i];
        ind[v]--;
        if (!ind[v]) q.push(v);
      }
    }
  }
  void buildGraph() {
    cnt = 0;
    memset(head, -1, sizeof(head));
    int u, v;
    scanf("%d", &u);
    for (int i = 1; i < n; i++) {
      scanf("%d", &v);
      addEdge(u, v);
      u = v;
    }
    scanf("%d", &u);
    for (int i = 1; i < n; i++) {
      scanf("%d", &v);
      addEdge(u, v);
      u = v;
    }
  }
  void solve() {
    buildGraph();
    for (int i = 1; i <= n; i++) {
      if (!dfn[i]) tarjan(i);
    }
    if (scc_cnt < k) {
      puts("NO");
      return;
    }
    get_scc();
    topoSort();
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%c", ans[belong[i]]);
    puts("");
  }
} scc;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  scc.init(n, k);
  scc.solve();
  return 0;
}

```