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 <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <set>
#include <map>
typedef long long ll;
using namespace std;

const int maxn = 1e5 + 79;
vector<vector<int> > s(maxn), nn(maxn);
vector<int> p(maxn), d(maxn, 0);
void dfs(int v = 0)
{
    nn[d[v]].push_back(v);
    for (int i : s[v])
    {
        d[i] = d[v] + 1;
        dfs(i);
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, x;
    cin >> n >> x;
    for (int i = 1; i < n; i++)
    {
        cin >> p[i];
        s[--p[i]].push_back(i);
    }
    //for (int i = 1; i < n; i++) cout << i + 1 << " " << p[i] + 1 << "\n";
    dfs();
    map<int, vector<int> > ip;
    for (int i = 0; i < n; i++) if (nn[i].size()) ip[nn[i].size()].push_back(i);
    vector<pair<int, int> > pv(n + 1, { -1, -1 }); // { z ktorej hodnoty sme to dostali, ktora uroven to bola }
    for (const pair<int, vector<int> >& i : ip)
    {
        for (int j = n; j >= 0; j--)
        {
            if (j && pv[j].first == -1) continue;
            for (int c = 1; c <= i.second.size() && j + c * i.first; c++)
            {
                if (pv[j + c * i.first].first == -1) pv[j + c * i.first] = { j, i.first };
            }
        }
    }
    if (!x || pv[x].first > -1)
    {
        cout << *max_element(d.begin(), d.end()) + 1 << "\n";
        string ans(n, 'b');
        while (x)
        {
            int nwx = pv[x].first, val = pv[x].second;
            int how_many = (x - nwx) / val;
            for (int i = 0; i < how_many; i++)
                for (int j : nn[ip[val][i]]) ans[j] = 'a';
            x = nwx;
        }
        cout << ans << "\n";
        return 0;
    }
    cout << *max_element(d.begin(), d.end()) + 2 << "\n";
    string ans(n, '?');
    int a = x, b = n - x;
    for (int i = 0; i < n; i++)
    {
        if (nn[i].size() <= a)
        {
            for (int j : nn[i]) ans[j] = 'a';
            a -= nn[i].size();
        }
        else if (nn[i].size() <= b)
        {
            for (int j : nn[i]) ans[j] = 'b';
            b -= nn[i].size();
        }
        else
        {
            bool takeb = (b > a);
            for (int j : nn[i])
            {
                if (!s[j].empty()) // non-leaf - chceme mu priradit to pismenko ktoreho mame viac
                {
                    if (takeb) ans[j] = 'b', b--;
                    else ans[j] = 'a', a--;
                }
            }
            for (int j : nn[i])
            {
                if (ans[j] == '?')
                {
                    if (b && takeb) ans[j] = 'b', b--;
                    else if (a && (!takeb)) ans[j] = 'a', a--;
                    else ans[j] = (a ? 'a' : 'b');
                }
            }
        }
    }
    cout << ans << "\n";
    return 0;
}