Problem: Kilani and Abd are neighbors for 3000 years, but then the day came and Kilani decided to move to another house. As a farewell gift, Kilani is going to challenge Abd with a problem written by their other neighbor with the same name Abd.

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The problem is:

You are given a connected tree rooted at node 1.

You should assign a character a or b to every node in the tree so that the total number of a's is equal to x and the total number of b's is equal to n - x.

Let's define a string for each node v of the tree as follows: 

  * if v is root then the string is just one character assigned to v: 
  * otherwise, let's take a string defined for the v's parent p_v and add to the end of it a character assigned to v. 



You should assign every node a character in a way that minimizes the number of distinct strings among the strings of all nodes.

Input

The first line contains two integers n and x (1 ≤ n ≤ 10^5; 0 ≤ x ≤ n) — the number of vertices in the tree the number of a's.

The second line contains n - 1 integers p_2, p_3, ..., p_{n} (1 ≤ p_i ≤ n; p_i ≠ i), where p_i is the parent of node i.

It is guaranteed that the input describes a connected tree.

Output

In the first line, print the minimum possible total number of distinct strings.

In the second line, print n characters, where all characters are either a or b and the i-th character is the character assigned to the i-th node.

Make sure that the total number of a's is equal to x and the total number of b's is equal to n - x.

If there is more than one answer you can print any of them.

Example

Input


9 3
1 2 2 4 4 4 3 1


Output


4
aabbbbbba

Note

The tree from the sample is shown below:

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The tree after assigning characters to every node (according to the output) is the following:

<image>

Strings for all nodes are the following: 

  * string of node 1 is: a
  * string of node 2 is: aa
  * string of node 3 is: aab
  * string of node 4 is: aab
  * string of node 5 is: aabb
  * string of node 6 is: aabb
  * string of node 7 is: aabb
  * string of node 8 is: aabb
  * string of node 9 is: aa



The set of unique strings is \{a, aa, aab, aabb\}, so the number of distinct strings is 4.

Solution (in CPP):

Solution:
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<climits>
#include<bitset>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int MAXN=1e5+11;
bitset<100001> f[501],ww; vector<int> vec[MAXN],V[MAXN],VV[MAXN],Vv[MAXN];
int N,siz[MAXN],cnt[MAXN],dep[MAXN],Num[MAXN],tot,K,vis[MAXN],Ans[MAXN];
pii sta[MAXN];
void dfs(int u,int fath){
    siz[u]=1,dep[u]=dep[fath]+1; cnt[dep[u]]++; VV[dep[u]].pb(u); Vv[dep[u]].pb(u);
    for(auto v:vec[u]) if(v!=fath) dfs(v,u),siz[u]+=siz[v]; return;
}
bool cmp(int x,int y){return siz[x]<siz[y];}
int main(){
    N=read(),K=read(); for(int i=2;i<=N;i++){int u=read(),v=i;vec[u].pb(v),vec[v].pb(u);}
    dfs(1,0); int Maxn=0; for(int i=1;i<=N;i++) Maxn=max(Maxn,dep[i]);
    for(int i=1;i<=Maxn;i++) V[cnt[i]].pb(i);
    for(int i=1;i<=N;i++) if(V[i].size()){++tot;sta[tot].fi=i,sta[tot].se=V[i].size();}
    f[0][0]=1;
    for(int i=1;i<=tot;i++){
        int v=sta[i].fi,c=sta[i].se;
        for(int j=0;j<=c;j++) f[i]|=f[i-1]<<(j*v);
    }
    if(f[tot][K]){
        printf("%d\n",Maxn); int now=K;
        for(int i=tot;i>=1;i--){
            int v=sta[i].fi,c=sta[i].se,ps=-1;
            for(int j=0;j<=c;j++) if(f[i-1][now-j*v]){ps=j;break;}
            for(int j=1;j<=ps;j++) vis[V[v][j-1]]=1;
            now-=ps*v;
        }
        for(int i=1;i<=N;i++) printf(vis[dep[i]]?"a":"b");printf("\n"); return 0;
    }
    printf("%d\n",Maxn+1); pii a=mp(0,K),b=mp(1,N-K);
    for(int i=1;i<=Maxn;i++){
        sort(VV[i].begin(),VV[i].end(),cmp); if(a.se<b.se) swap(a,b);
        while(VV[i].size()&&a.se) Ans[VV[i].back()]=a.fi,a.se--,VV[i].pop_back();
        while(VV[i].size()&&b.se) Ans[VV[i].back()]=b.fi,b.se--,VV[i].pop_back();
    }
    for(int i=1;i<=N;i++) printf(!Ans[i]?"a":"b"); printf("\n"); return 0;
}