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;

const int N = 100007, M = 500007, INF = 0x3f3f3f3f;

typedef long long ll;
typedef int itn;

int n, m, t, x;
int fa[N];
int deep[N];
int maxx;
int num[N];
int cnt;
int lson[N];
int head[N], ver[M], edge[M], nex[M], tot;
bitset<N> f[2007];
int val[N];
unordered_map<int, int> vis;
bool ok[N];
vector<int> v[N];

void add(int x, itn y)
{
    ver[tot] = y;
    nex[tot] = head[x];
    head[x] = tot++;
}

void dfs1(int x)
{
    deep[x] = deep[fa[x]] + 1;
    num[deep[x]]++;
    maxx = max(maxx, deep[x]);
    if (head[x] == -1)
        lson[deep[x]]++;
    for (int i = head[x]; ~i; i = nex[i])
    {
        int y = ver[i];
        dfs1(y);
    }
}

void dfs2(itn x, int t)
{
    if (x == 0)
        return;
    for (int i = 0; i < (int)v[x].size(); ++i)
    {
        if (val[x] > t || f[x - 1][t])
            break;
        t -= val[x], ok[v[x][i]] = true;
    }
    dfs2(x - 1, t);
}

void init()
{
    memset(head, -1, sizeof head);
    tot = 0;
}

void solve()
{
    init();
    scanf("%d%d", &n, &x);
    for (int i = 2; i <= n; ++i)
    {
        scanf("%d", &fa[i]);
        add(fa[i], i);
    }
    dfs1(1);
    for (int i = 1; i <= maxx; ++i)
    {
        if (vis[num[i]])
        {
            v[vis[num[i]]].push_back(i);
        }
        else
        {
            vis[num[i]] = ++cnt;
            val[cnt] = num[i];
            v[cnt].push_back(i);
        }
    }

    f[0][0] = 1;

    for (int i = 1; i <= cnt; ++i)
    {
        f[i] = f[i - 1];
        itn Size = v[i].size();
        for (int j = 1; j <= Size; j <<= 1)
        {
            Size -= j;
            f[i] |= (f[i] << (j * val[i]));
        }
        if (Size > 0)
            f[i] |= f[i] << (Size * val[i]);
    }

    if (f[cnt][x])
    {
        printf("%d\n", maxx);
        dfs2(cnt, x);
        for (int i = 1; i <= n; ++i)
        {
            putchar(ok[deep[i]] ? 'a' : 'b');
        }
    }
    else
    {
        int res = INF;
        for (int i = x; i >= 0; --i)
        {
            if (f[cnt][i])
            {
                res = i;
                break;
            }
        }
        dfs2(cnt, res);
        int pos = -1;
        for (int i = 1; i <= maxx; ++i)
        {
            if (!ok[i] && lson[i] >= x - res)
            {
                pos = i;
                break;
            }
        }

        printf("%d\n", maxx + 1);

        for (int i = 1; i <= n; ++i)
        {
            if (deep[i] == pos && head[i] == -1)
            {
                if (res == x)
                {
                    putchar('b');
                }
                else
                    putchar('a'), ++res;
            }
            else
            {
                putchar(ok[deep[i]] ? 'a' : 'b');
            }
        }
    }
    return;
}

itn main()
{
    solve();
    return 0;
}