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;
bool db[100000][500];
int maxHeight = 0;


int main() {
    int n, x;
    cin >> n >> x;
    vector<int> T[n + 1];
    for (int i = 2; i <= n; i++) {
        int node;
        cin >> node;
        T[node].push_back(i);
    }
    //Determine wether the path exists or not
    vector<int> heights;
    vector<int> openList(1, 1);
    vector<vector<int>> heightNodes(n);
    heightNodes[0].push_back(1);
    int height = 1;
    while (!openList.empty()) {
        maxHeight++;
        vector<int> nodes;
        for (int node : openList)
            for (int child: T[node]){
                nodes.push_back(child);
                heightNodes[height].push_back(child);
            }
        height++;
        heights.push_back(openList.size());
        openList = nodes;
    }
    sort(heights.begin(), heights.end());
    vector<pair<int, int>> heightCount;
    for (int i = 0; i < heights.size(); i++) {
        if (i == 0 or heightCount.back().first != heights[i])
            heightCount.push_back(pair<int, int>(heights[i], 1));
        else
            heightCount.back().second++;
    }
    db[0][heightCount.size()] = true;
    for(int i = heightCount.size() - 1 ; i >= 0 ; i--){
        int h = heightCount[i].first;
        int f = heightCount[i].second;
        vector<int> last(h,-1);
        for(int j = 0 ; j <= x ; j++){
            if(db[j][i+1]){
                db[j][i] = true;
                last[j%h] = j;
            }
            else if( last[j%h] == -1)
                db[j][i] = false;
            else if((j - last[j%h])/h <= f)
                db[j][i] = true;
            else
                db[j][i] = false;
        }
    }
    if (db[x][0]) {

        //if the path exists
        vector<int> aCount(n + 1);
        vector<char> labels(n + 1);
        int h = heightCount[0].first;
        int sum = x;
        int index = 0;
        while (sum > 0) {
            if (db[sum][ index + 1]) {
                index++;
                h = heightCount[index].first;

            } else {
                sum -= h;
                aCount[h]++;
            }
        }
        for(int i = 0 ; i < maxHeight; i++){
            h = heightNodes[i].size();
            if(aCount[h] > 0){
                aCount[h] --;
                for(int node : heightNodes[i]){
                    labels[node] = 'a';
                }
            }


            else{
                for(int node : heightNodes[i]){
                    labels[node] = 'b';
                }
            }
        }

        cout << maxHeight << endl;
        for (int i = 1; i <= n; i++)
            cout << labels[i] ;
    } else {

        int aCount = x;
        int bCount = n - x;
        char labels[n + 1];
        vector<int> openList(1, 1);
        while (!openList.empty()) {
            vector<int> leaves;
            vector<int> nonLeaves;
            for (int node : openList) {
                if (T[node].empty())
                    leaves.push_back(node);
                else
                    nonLeaves.push_back(node);
            }
            if (aCount > bCount) {
                for (int node : nonLeaves) {
                    labels[node] = 'a';
                    aCount--;
                }
                for (int node : leaves) {
                    if (aCount > 0) {
                        labels[node] = 'a';
                        aCount--;
                    } else {
                        labels[node] = 'b';
                        bCount--;
                    }
                }
            }
            else{
                for (int node : nonLeaves) {
                    labels[node] = 'b';
                    bCount--;
                }
                for (int node : leaves) {
                    if (bCount > 0) {
                        labels[node] = 'b';
                        bCount--;
                    } else {
                        labels[node] = 'a';
                        aCount--;
                    }
                }
            }
            vector<int> newNodes;
            for(int node : openList){
                for(int child : T[node])
                    newNodes.push_back(child);
            }
            openList = newNodes;



        }
        cout << maxHeight+1 << endl;
        for(int i =1 ; i<=n ;i++)
            cout << labels[i];
    }

}