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.

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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 <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
#define Ri register
typedef long long ll;

const int N = 1e5 + 100, M = 500;

int n, f[N], son[N], dep[N], A, B, maxdep;
int ansnum, ls[N], tot, fre[N], dp[M][N];
char ans[N];
vector<int> ed[N], d[N];
int cnt[N];

void dfs(int u) {
    dep[u] = dep[f[u]]+ 1;
    d[dep[u]].push_back(u);
    maxdep = max(maxdep, dep[u]);
    for (auto v : ed[u]) {
        if (v == f[u]) continue;
        dfs(v);
    }
}

int Find(int x) {
    return lower_bound(ls + 1, ls + 1 + tot, x) - ls;
}

int cmp(int a, int b) {
    return son[a] > son[b];
}

void findcnt(int layer, int cur) {
    if (layer == 0) return;
    cnt[layer] = dp[layer][cur];
    findcnt(layer-1, cur - cnt[layer] * ls[layer]);
}

void solve(int pos) {
    int num = d[pos].size();
    if (min(A, B) >= num) {
        if (A > B) {
            for (auto i : d[pos])
                ans[i] = 'b';
            B -= num;
        }
        else {
            for (auto i : d[pos])
                ans[i] = 'a';
            A -= num;
        }
    }
    else if (max(A, B) >= num) {
        if (A >= num) {
            for (auto i : d[pos])
                ans[i] = 'a';
            A -= num;
        }
        else {
            for (auto i : d[pos])
                ans[i] = 'b';
            B -= num;
        }
    }
    else {
        ansnum++;
        if (A > B) {
            for (auto i : d[pos]) {
                if (A) { ans[i] = 'a'; A--; }
                else { ans[i] = 'b'; B--; }
            }
        }
        else {
            for (auto i : d[pos]) {
                if (B) {ans[i] = 'b'; B--;}
                else {ans[i] = 'a'; A--;}
            }
        }
    }
}

int main() {
    scanf("%d%d", &n, &A);
    B = n - A;
    for (int i = 2; i <= n; ++i) {
        scanf("%d", &f[i]);
        son[f[i]]++;
        ed[i].push_back(f[i]);
        ed[f[i]].push_back(i);
    }
    dfs(1);
    ansnum = maxdep;
    for (int i = 1; i <= maxdep; ++i)
        sort(d[i].begin(), d[i].end(), cmp);
    for (int i = 1; i <= maxdep; ++i)
        ls[i] = d[i].size();
    tot = maxdep;
    sort(ls + 1, ls + 1 + tot);
    tot = unique(ls + 1, ls + 1 + tot) - ls - 1;
    for (int i = 1; i <= maxdep; ++i)
        fre[Find(d[i].size())]++;

    memset(dp, -1, sizeof(dp));
    dp[0][0] = 1;    
    for (int i = 1; i <= tot; ++i) {
        for (int j = 0; j <= n; ++j) {
            if (dp[i - 1][j] >= 0) 
                dp[i][j] = 0;
            else if (j - ls[i] >= 0 && dp[i][j - ls[i]] >= 0 && dp[i][j - ls[i]] < fre[i])
                dp[i][j] = dp[i][j - ls[i]] + 1;
        }
    }
    if (dp[tot][A] >= 0) {
        findcnt(tot, A);
        for (int i = 1; i <= maxdep; ++i) {
            int x = Find(d[i].size());
            if (cnt[x]) {
                cnt[x]--;
                for (auto j : d[i])
                    ans[j] = 'a';
            }
            else {
                for (auto j : d[i])
                    ans[j] = 'b';
            }
        }
        printf("%d\n", maxdep);
        printf("%s\n", ans + 1);
        return 0;
    }
    for (int i = 1; i <= maxdep; ++i) {
        solve(i);
    }    
    printf("%d\n", ansnum);
    printf("%s\n", ans + 1);
    return 0;
}