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.

<image>

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<vector>
//#include<queue>
//#include<sstream>
//#include<cmath>
//#include<set>
//#include<cstdio>
//#include<algorithm>
//#include<map>
#include<bits/stdc++.h>
#define mp make_pair
#define pii pair<int,int>
#pragma GCC optimize ("O2")
#define pdd pair<double,double>
#define pll pair<ll,ll>
#define mst(a,b) memset(a,b,sizeof(a))
#pragma comment(linker, "/STACK:102400000,102400000")
//g++ strlen.cpp -o a.exe -Wl,--stack,536870912
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<double> vec;
typedef vector<vec> mat;
const double eps = 1e-8;
const int maxn = 1e5+9;
const double pi=acos(-1.0);
const int inf = 2e9;
const ll mod = 1e9+7;
vector<int> g[maxn],level[maxn],op[maxn];
int n,x,dp[500][maxn],pos,tag[maxn],num;
char ans[maxn];
pii pq[maxn];
void dfs(int p,int jk){
    level[jk].push_back(p);
    tag[p]=jk;
    pos=max(pos,jk);
    for(auto i:g[p]) dfs(i,jk+1);
}
void init(){
    pos=0;dfs(1,0);
    ++pos;
    for(int i=0;i<pos;++i){
        op[level[i].size()].push_back(i);
    }
    num=0;
    for(int i=1;i<=n;++i) if(op[i].size()) pq[num++]=mp(i,op[i].size());
//    for(int i=0;i<num;++i) cout<<pq[i].first<<","<<pq[i].second<<endl;
    mst(dp[0],-1);
    for(int i=0;i<=pq[0].second;++i) dp[0][i*pq[0].first]=pq[0].second-i;
    for(int i=1;i<num;++i){
        for(int j=0;j<=n;++j){
            if(dp[i-1][j]>=0) dp[i][j]=pq[i].second;
            else dp[i][j]=-1;
        }
        for(int j=0;j<=n-pq[i].first;++j){
            if(dp[i][j]>0) dp[i][j+pq[i].first]=max(dp[i][j+pq[i].first],dp[i][j]-1);
        }
//        for(int j=0;j<=n;++j) cout<<dp[i][j]<<" ";
//        cout<<endl;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>x;
    for(int i=2;i<=n;++i){
        int tmp;
        cin>>tmp;
        g[tmp].push_back(i);
    }
    init();
    if(dp[num-1][x]>=0){
        cout<<pos<<"\n";
        int jk=x;
        for(int i=num-1;i>=0;--i){
            if(dp[i][jk]<pq[i].second){
                for(int j=0;j<pq[i].second-dp[i][jk];++j){
                    vector<int> &tmp=level[op[pq[i].first][j]];
                    for(auto k:tmp) ans[k]='a';
                }
                for(int j=pq[i].second-dp[i][jk];j<pq[i].second;++j){
                    vector<int> &tmp=level[op[pq[i].first][j]];
                    for(auto k:tmp) ans[k]='b';
                }
                jk-=pq[i].first*(pq[i].second-dp[i][jk]);
            }
            else{
                for(int j=0;j<pq[i].second;++j){
                    vector<int> &tmp=level[op[pq[i].first][j]];
                    for(auto k:tmp) ans[k]='b';
                }
            }
        }
    }
    else{
        cout<<pos+1<<"\n";
        int now=x,le=n;
        for(int i=0;i<pos;++i){
            if(level[i].size()<=now){
                now-=level[i].size();
                for(auto j:level[i]) ans[j]='a';
            }
            else if(level[i].size()<=le-now){
                for(auto j:level[i]) ans[j]='b';
            }
            else{
                char tmp='a',jk='b';
                int kl=now,io=le-now;
                if(le-now>now) swap(tmp,jk),swap(kl,io);
                for(auto j:level[i]){
                    if(g[j].size()){
                        ans[j]=tmp;
                        --kl;
                    }
                }
                for(auto j:level[i]){
                    if(!g[j].size()){
                        if(kl){
                            --kl;
                            ans[j]=tmp;
                        }
                        else{
                            --io;
                            ans[j]=jk;
                        }
                    }
                }
                if(tmp=='a') now=kl;
                else now=io;
            }
            le-=level[i].size();
        }
    }
    for(int i=1;i<=n;++i) cout<<ans[i];
    return 0;
}