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 BSZ = 500;

int N, X;
vector<int> adj[101010];
vector<int> ds[101010];
vector<int> vs[101010];
int dep[101010];
int D[101010];
int path[101010];
int cnt[101010];
bool chk[101010];
char ans[101010];
int mx;
int id;

map<int, int> mp;

void dfs(int v) {
	mx = max(mx, dep[v]);
	cnt[dep[v]]++;
	vs[dep[v]].push_back(v);
	for(int i : adj[v]) {
		dep[i] = dep[v] + 1;
		dfs(i);
	}
}

int main() {
	scanf("%d%d", &N, &X);

	for(int i = 2; i <= N; i++) {
		int p; scanf("%d", &p);
		adj[p].push_back(i);
	}

	dep[1] = 1; dfs(1);

	D[0] = 1;
	for(int i = 1; i <= mx; i++) {
		mp[cnt[i]]++;
		ds[cnt[i]].push_back(i);
	}
	for(auto& i : mp) {
		for(int j = 0; j < i.first; j++) {
			int lft = 0;
			for(int k = j; k <= N; k += i.first) {
				if(D[k]) lft = i.second;
				//else if(lft && k >= i.first && D[k - i.first]) {
				else if(lft) {
					D[k] = 1; lft--;
					path[k] = ds[i.first][lft];
				}
			}
		}
	}

	if(D[X]) {
		printf("%d\n", mx);
		int now = X;
		while(now) {
			chk[path[now]] = true;
			now -= cnt[path[now]];
		}
		for(int i = 1; i <= N; i++) printf("%c", chk[dep[i]] ? 'a' : 'b'); puts("");
		return 0;
	}

	int A = X, B = N - X;
	for(int i = 1; i <= mx; i++) {
		if(A >= cnt[i]) {
			for(int j : vs[i]) ans[j] = 'a';
			A -= cnt[i];
		}
		else if(B >= cnt[i]) {
			for(int j : vs[i]) ans[j] = 'b';
			B -= cnt[i];
		}
		else if(A > B) {
			for(int j : vs[i]) {
				if(adj[j].size()) { A--; ans[j] = 'a'; }
			}
			for(int j : vs[i]) {
				if(adj[j].size()) continue;
				if(A) { A--; ans[j] = 'a'; }
				else { B--; ans[j] = 'b'; }
			}
		}
		else {
			for(int j : vs[i]) {
				if(adj[j].size()) { B--; ans[j] = 'b'; }
			}
			for(int j : vs[i]) {
				if(adj[j].size()) continue;
				if(B) { B--; ans[j] = 'b'; }
				else { A--; ans[j] = 'a'; }
			}
		}
	}

	printf("%d\n", mx + 1);
	printf("%s\n", ans + 1);

	return 0;
}