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 <bits/stdc++.h>
using namespace std;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl





using ll = long long;


const int maxn = 1e5 + 5;
const int SQ = 500; //sqrt(1e5)=316

int n, x;

pair<int, char> a,b;

int mx;


vector<int> g[maxn];
vector<int> nodes[maxn]; // by depth
int siz[maxn];

int dfs(int at, int dep) {
    mx = max(mx, dep);
    nodes[dep].push_back(at);
    siz[at] = 1;
    for (int to: g[at]) {
	siz[at] += dfs(to, dep+1);
    }
    return siz[at];
}


bool dp[SQ][maxn];


vector<pair<int,int>> v, v2;
int f[maxn];

void build(int i, int have) {
    if (i == (int)v2.size()) {
	return;
    }
    while (!dp[i+1][have]) {
	f[v2[i].first]++;
	have -= v2[i].first;
    }
    build(i+1, have);
}


int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    cin>>n>>x;

    a = {x, 'a'};
    b = {n-x, 'b'};
    if (a>b) {
	swap(a,b);
    }

    for (int i=2; i<=n; i++) {
	int p;
	cin>>p;
	g[p].push_back(i);
    }


    dfs(1, 0);



    for (int i=0; i<=mx; i++) {
	v.push_back({int(nodes[i].size()), i}); // size, dep
    }
    sort(v.begin(), v.end());
    for (int i=0; i<(int)v.size(); i++) {
	if (i==0 || v[i].first != v[i-1].first) {
	    v2.push_back({v[i].first, 1}); //size, freq
	} else {
	    v2.back().second++;
	}
    }

    int N = v2.size();
    dp[N][0] = true;
    for (int i=N-1; i>=0; i--) {
	int val = v2[i].first;
	int freq = v2[i].second;
	vector<int> last(val, -1);
	for (int j=0; j<=a.first; j++) {
	    if (dp[i+1][j]) {
		last[j%val] = j; //largest k
	    }
	    if (last[j%val]==-1 || (j-last[j%val])/val > freq) {
		dp[i][j] = false;
	    } else {
		dp[i][j] = true;
	    }
	}
    }

    if (dp[0][a.first]) {
	build(0, a.first);
	string ans(n, b.second);
	for (int i=0; i<=mx; i++) {
	    if (f[(int)nodes[i].size()] == 0) {
		continue;
	    }
	    f[(int)nodes[i].size()]--;
	    for (int x: nodes[i]) {
		ans[x-1] = a.second;
	    }
	}

	cout<<mx+1<<endl;
	cout<<ans<<endl;
	exit(0);
    }

    cout<<mx+2<<endl;
    string ans(n, '.');
    
    for (int i=0; i<=mx; i++) {
	sort(nodes[i].begin(), nodes[i].end(), [&](int u, int v) {
	    return siz[u] < siz[v];
	});


	if (a.first < b.first) {
	    swap(a,b);
	}

	while (!nodes[i].empty() && a.first > 0) {
	    ans[nodes[i].back()-1] = a.second;
	    nodes[i].pop_back();
	    a.first--;
	}

	while (!nodes[i].empty() && b.first > 0) {
	    ans[nodes[i].back()-1] = b.second;
	    nodes[i].pop_back();
	    b.first--;
	}
    }

    cout<<ans<<endl;
    
    
    return 0;
}